Not able to get php parsed data in framework7 through ajax call
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}
1
function pendingorder(){
app.request({
type:"POST",
url: "pages/getpeningorer.php",
dataType: 'json',
cache: false,
success:function(data) {
console.log(data);
var result = $.parseJSON(data);
$.each(result, function(key, value){
$.each(value, function(k, v){
if(k === "order_id"){
$("#pendingtable >tbody:last").append(
$('<tr>').append(
$('<td>').append(v)
.append(
$('</td>').append(
$('</tr>')
)
)
)
);
}
if(k === "product_id"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
if(k === "status"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
if(k === "remark"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
if(k === "postingDate"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
});
});
console.log(data);}
});
console.log('execute success');
}
I AM trying to call ajax through function....But not working. In a similar way, I post data it is working.
PHP code:
enter code here
$conn = new mysqli($host, $dbUsername, $dbPassword, $dbname);
if (mysqli_connect_error()) {
die('Connect Error('. mysqli_connect_errno().')'. mysqli_connect_error());
} else {
$sql = "SELECT order_id,product_id,status,remark,postingDate FROM order_track_history where status='In process'";
$result = $conn->query($sql);
}
$Pdata = array();
while ($row = mysql_fetch_array($result)) {
$picture = array(
"order_id" => $row['order_id'],
"product_id" => $row['product_id'],
"status" => $row['status'],
"remark" => $row['remark'],
"postingDate" => $row['postingDate']
);
$Pdata = $picture;
}`enter code here`
echo json_encode($Pdata);
Here I am sending my data to ajax calls in JSON format. But not able to see data at HTML page.
jquery mysql ajax html-framework-7
edited Nov 29 '18 at 5:33
Suresh Kamrushi
11k95772
asked Nov 29 '18 at 5:32
MANDAR MAHADEOKARMANDAR MAHADEOKAR
63
add a comment |
1
function pendingorder(){
app.request({
type:"POST",
url: "pages/getpeningorer.php",
dataType: 'json',
cache: false,
success:function(data) {
console.log(data);
var result = $.parseJSON(data);
$.each(result, function(key, value){
$.each(value, function(k, v){
if(k === "order_id"){
$("#pendingtable >tbody:last").append(
$('<tr>').append(
$('<td>').append(v)
.append(
$('</td>').append(
$('</tr>')
)
)
)
);
}
if(k === "product_id"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
if(k === "status"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
if(k === "remark"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
if(k === "postingDate"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
});
});
console.log(data);}
});
console.log('execute success');
}
I AM trying to call ajax through function....But not working. In a similar way, I post data it is working.
PHP code:
enter code here
$conn = new mysqli($host, $dbUsername, $dbPassword, $dbname);
if (mysqli_connect_error()) {
die('Connect Error('. mysqli_connect_errno().')'. mysqli_connect_error());
} else {
$sql = "SELECT order_id,product_id,status,remark,postingDate FROM order_track_history where status='In process'";
$result = $conn->query($sql);
}
$Pdata = array();
while ($row = mysql_fetch_array($result)) {
$picture = array(
"order_id" => $row['order_id'],
"product_id" => $row['product_id'],
"status" => $row['status'],
"remark" => $row['remark'],
"postingDate" => $row['postingDate']
);
$Pdata = $picture;
}`enter code here`
echo json_encode($Pdata);
Here I am sending my data to ajax calls in JSON format. But not able to see data at HTML page.
jquery mysql ajax html-framework-7
edited Nov 29 '18 at 5:33
Suresh Kamrushi
11k95772
asked Nov 29 '18 at 5:32
MANDAR MAHADEOKARMANDAR MAHADEOKAR
63
have you tried console.log the data return by your php script?
– Buddy
Nov 29 '18 at 5:37
not getting data also in console
– MANDAR MAHADEOKAR
Nov 29 '18 at 5:38
if you access this pagepages/getpeningorer.phpdirect from browser, what is result will you got?
– Anees Hikmat Abu Hmiad
Nov 29 '18 at 14:44
-- also make sure you are enable error in php.ini or set this code in your php fileerror_reporting(E_ALL); ini_set('display_errors', 1);
– Anees Hikmat Abu Hmiad
Nov 29 '18 at 14:49
work after setting var result = $.parseJSON(data); to var result = data;
– MANDAR MAHADEOKAR
Dec 3 '18 at 8:57
add a comment |
1
1
1
function pendingorder(){
app.request({
type:"POST",
url: "pages/getpeningorer.php",
dataType: 'json',
cache: false,
success:function(data) {
console.log(data);
var result = $.parseJSON(data);
$.each(result, function(key, value){
$.each(value, function(k, v){
if(k === "order_id"){
$("#pendingtable >tbody:last").append(
$('<tr>').append(
$('<td>').append(v)
.append(
$('</td>').append(
$('</tr>')
)
)
)
);
}
if(k === "product_id"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
if(k === "status"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
if(k === "remark"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
if(k === "postingDate"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
});
});
console.log(data);}
});
console.log('execute success');
}
I AM trying to call ajax through function....But not working. In a similar way, I post data it is working.
PHP code:
enter code here
$conn = new mysqli($host, $dbUsername, $dbPassword, $dbname);
if (mysqli_connect_error()) {
die('Connect Error('. mysqli_connect_errno().')'. mysqli_connect_error());
} else {
$sql = "SELECT order_id,product_id,status,remark,postingDate FROM order_track_history where status='In process'";
$result = $conn->query($sql);
}
$Pdata = array();
while ($row = mysql_fetch_array($result)) {
$picture = array(
"order_id" => $row['order_id'],
"product_id" => $row['product_id'],
"status" => $row['status'],
"remark" => $row['remark'],
"postingDate" => $row['postingDate']
);
$Pdata = $picture;
}`enter code here`
echo json_encode($Pdata);
Here I am sending my data to ajax calls in JSON format. But not able to see data at HTML page.
jquery mysql ajax html-framework-7
edited Nov 29 '18 at 5:33
Suresh Kamrushi
11k95772
asked Nov 29 '18 at 5:32
MANDAR MAHADEOKARMANDAR MAHADEOKAR
63
function pendingorder(){
app.request({
type:"POST",
url: "pages/getpeningorer.php",
dataType: 'json',
cache: false,
success:function(data) {
console.log(data);
var result = $.parseJSON(data);
$.each(result, function(key, value){
$.each(value, function(k, v){
if(k === "order_id"){
$("#pendingtable >tbody:last").append(
$('<tr>').append(
$('<td>').append(v)
.append(
$('</td>').append(
$('</tr>')
)
)
)
);
}
if(k === "product_id"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
if(k === "status"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
if(k === "remark"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
if(k === "postingDate"){
$("#demoTable >tbody >tr:last").append(
$('<td>').append(v)
.append(
$('</td>')
)
);
}
});
});
console.log(data);}
});
console.log('execute success');
}
I AM trying to call ajax through function....But not working. In a similar way, I post data it is working.
PHP code:
enter code here
$conn = new mysqli($host, $dbUsername, $dbPassword, $dbname);
if (mysqli_connect_error()) {
die('Connect Error('. mysqli_connect_errno().')'. mysqli_connect_error());
} else {
$sql = "SELECT order_id,product_id,status,remark,postingDate FROM order_track_history where status='In process'";
$result = $conn->query($sql);
}
$Pdata = array();
while ($row = mysql_fetch_array($result)) {
$picture = array(
"order_id" => $row['order_id'],
"product_id" => $row['product_id'],
"status" => $row['status'],
"remark" => $row['remark'],
"postingDate" => $row['postingDate']
);
$Pdata = $picture;
}`enter code here`
echo json_encode($Pdata);
Here I am sending my data to ajax calls in JSON format. But not able to see data at HTML page.
jquery mysql ajax html-framework-7
jquery mysql ajax html-framework-7
edited Nov 29 '18 at 5:33
Suresh Kamrushi
11k95772
asked Nov 29 '18 at 5:32
MANDAR MAHADEOKARMANDAR MAHADEOKAR
63
edited Nov 29 '18 at 5:33
Suresh Kamrushi
11k95772
asked Nov 29 '18 at 5:32
MANDAR MAHADEOKARMANDAR MAHADEOKAR
63
edited Nov 29 '18 at 5:33
Suresh Kamrushi
11k95772
edited Nov 29 '18 at 5:33
Suresh Kamrushi
11k95772
edited Nov 29 '18 at 5:33
Suresh Kamrushi
11k95772
11k95772
asked Nov 29 '18 at 5:32
MANDAR MAHADEOKARMANDAR MAHADEOKAR
63
asked Nov 29 '18 at 5:32
MANDAR MAHADEOKARMANDAR MAHADEOKAR
63
asked Nov 29 '18 at 5:32
MANDAR MAHADEOKARMANDAR MAHADEOKAR
63
63
have you tried console.log the data return by your php script?
– Buddy
Nov 29 '18 at 5:37
not getting data also in console
– MANDAR MAHADEOKAR
Nov 29 '18 at 5:38
if you access this pagepages/getpeningorer.phpdirect from browser, what is result will you got?
– Anees Hikmat Abu Hmiad
Nov 29 '18 at 14:44
-- also make sure you are enable error in php.ini or set this code in your php fileerror_reporting(E_ALL); ini_set('display_errors', 1);
– Anees Hikmat Abu Hmiad
Nov 29 '18 at 14:49
work after setting var result = $.parseJSON(data); to var result = data;
– MANDAR MAHADEOKAR
Dec 3 '18 at 8:57
add a comment |
have you tried console.log the data return by your php script?
– Buddy
Nov 29 '18 at 5:37
not getting data also in console
– MANDAR MAHADEOKAR
Nov 29 '18 at 5:38
if you access this pagepages/getpeningorer.phpdirect from browser, what is result will you got?
– Anees Hikmat Abu Hmiad
Nov 29 '18 at 14:44
-- also make sure you are enable error in php.ini or set this code in your php fileerror_reporting(E_ALL); ini_set('display_errors', 1);
– Anees Hikmat Abu Hmiad
Nov 29 '18 at 14:49
work after setting var result = $.parseJSON(data); to var result = data;
– MANDAR MAHADEOKAR
Dec 3 '18 at 8:57
have you tried console.log the data return by your php script?
– Buddy
Nov 29 '18 at 5:37
have you tried console.log the data return by your php script?
– Buddy
Nov 29 '18 at 5:37
not getting data also in console
– MANDAR MAHADEOKAR
Nov 29 '18 at 5:38
not getting data also in console
– MANDAR MAHADEOKAR
Nov 29 '18 at 5:38
if you access this page
pages/getpeningorer.php direct from browser, what is result will you got?– Anees Hikmat Abu Hmiad
Nov 29 '18 at 14:44
if you access this page
pages/getpeningorer.php direct from browser, what is result will you got?– Anees Hikmat Abu Hmiad
Nov 29 '18 at 14:44
-- also make sure you are enable error in php.ini or set this code in your php file
error_reporting(E_ALL); ini_set('display_errors', 1);– Anees Hikmat Abu Hmiad
Nov 29 '18 at 14:49
-- also make sure you are enable error in php.ini or set this code in your php file
error_reporting(E_ALL); ini_set('display_errors', 1);– Anees Hikmat Abu Hmiad
Nov 29 '18 at 14:49
work after setting var result = $.parseJSON(data); to var result = data;
– MANDAR MAHADEOKAR
Dec 3 '18 at 8:57
work after setting var result = $.parseJSON(data); to var result = data;
– MANDAR MAHADEOKAR
Dec 3 '18 at 8:57
add a comment |
1 Answer
1
active
oldest
votes
0
it seems like there's something wrong with your php codes, try this code instead
$row = mysql_fetch_array($result);
foreach($row as $r) {
$picture = array(
"order_id" => $r['order_id'],
"product_id" => $r['product_id'],
"status" => $r['status'],
"remark" => $r['remark'],
"postingDate" => $r['postingDate']
);
}
answered Nov 29 '18 at 5:46
BuddyBuddy
767
still Not working
– MANDAR MAHADEOKAR
Nov 29 '18 at 6:08
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
it seems like there's something wrong with your php codes, try this code instead
$row = mysql_fetch_array($result);
foreach($row as $r) {
$picture = array(
"order_id" => $r['order_id'],
"product_id" => $r['product_id'],
"status" => $r['status'],
"remark" => $r['remark'],
"postingDate" => $r['postingDate']
);
}
answered Nov 29 '18 at 5:46
BuddyBuddy
767
still Not working
– MANDAR MAHADEOKAR
Nov 29 '18 at 6:08
add a comment |
0
it seems like there's something wrong with your php codes, try this code instead
$row = mysql_fetch_array($result);
foreach($row as $r) {
$picture = array(
"order_id" => $r['order_id'],
"product_id" => $r['product_id'],
"status" => $r['status'],
"remark" => $r['remark'],
"postingDate" => $r['postingDate']
);
}
answered Nov 29 '18 at 5:46
BuddyBuddy
767
still Not working
– MANDAR MAHADEOKAR
Nov 29 '18 at 6:08
add a comment |
0
0
0
it seems like there's something wrong with your php codes, try this code instead
$row = mysql_fetch_array($result);
foreach($row as $r) {
$picture = array(
"order_id" => $r['order_id'],
"product_id" => $r['product_id'],
"status" => $r['status'],
"remark" => $r['remark'],
"postingDate" => $r['postingDate']
);
}
answered Nov 29 '18 at 5:46
BuddyBuddy
767
it seems like there's something wrong with your php codes, try this code instead
$row = mysql_fetch_array($result);
foreach($row as $r) {
$picture = array(
"order_id" => $r['order_id'],
"product_id" => $r['product_id'],
"status" => $r['status'],
"remark" => $r['remark'],
"postingDate" => $r['postingDate']
);
}
answered Nov 29 '18 at 5:46
BuddyBuddy
767
answered Nov 29 '18 at 5:46
BuddyBuddy
767
answered Nov 29 '18 at 5:46
BuddyBuddy
767
answered Nov 29 '18 at 5:46
BuddyBuddy
767
767
still Not working
– MANDAR MAHADEOKAR
Nov 29 '18 at 6:08
add a comment |
still Not working
– MANDAR MAHADEOKAR
Nov 29 '18 at 6:08
still Not working
– MANDAR MAHADEOKAR
Nov 29 '18 at 6:08
still Not working
– MANDAR MAHADEOKAR
Nov 29 '18 at 6:08
add a comment |
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have you tried console.log the data return by your php script?
– Buddy
Nov 29 '18 at 5:37
not getting data also in console
– MANDAR MAHADEOKAR
Nov 29 '18 at 5:38
if you access this page
pages/getpeningorer.phpdirect from browser, what is result will you got?– Anees Hikmat Abu Hmiad
Nov 29 '18 at 14:44
-- also make sure you are enable error in php.ini or set this code in your php file
error_reporting(E_ALL); ini_set('display_errors', 1);– Anees Hikmat Abu Hmiad
Nov 29 '18 at 14:49
work after setting var result = $.parseJSON(data); to var result = data;
– MANDAR MAHADEOKAR
Dec 3 '18 at 8:57