I am using ajax call for sending value to php file and get response back into first page,(work fine), but when i am getting the result in loop form this response nothing will be return on first page.
My first page:
$.ajax({
url: "action.php",
method: "post",
data: {
'trader': selected_trader
},
dataType: 'JSON',
success: function (response) {
$('#date').html(response['date']);
$('#symbol').html(response['symbol']);
$('#status').html(response['status']);
$('#alert').html(response['alert']);
$('#company_name').html(response['company_name']);
}
})
My 2nd page:
if (isset($_POST['trader'])) {
$trader = $_POST['trader'];
$sql = "Select * from current_trader where status='$trader' ";
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_assoc($result)) {
$data["date"] = date('Y:m:d', strtotime($row['date']));
$data["symbol"] = $row['symbol'];
$data["company_name"] = $row['company_name'];
$data["alert"] = $row['alert'];
$data["status"] = $row['status'];
echo json_encode($data);
}
}
Related
I have drop down Select box as follows
<?php
$sql = "SELECT scheduleName FROM schedule";
$result = mysqli_query($link,$sql);
echo "<select name='schedule' id='schedule'>";
echo "<option value=''>-- Select Schedule --</option>";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['scheduleName'] . "'>" . $row['scheduleName'] . "</option>";
}
echo "</select>";
?>
And I have a file called processClg.php as follows
<?php
include "config.php";
if ($_POST['type']=='POST')
{
$qry = "SELECT * FROM schedule WHERE scheduleName LIKE 'Row id of drop down selection'";
$res = mysqli_query($link,$qry);
}
?>
How can I call processClg.php file on $("#schedule").change(function ());by assigning Row id of drop down selection as where condition.
Update
Am getting Response from processClg.Php as follows
[{"id":"2","scheduleName":"shanth","subject":"Patho","university":"Dali","facultyName":"Dr","scheduleStartDate":"2015-06-05","scheduleEndDate":"2015-06-09"}]
How to assign response values from ajax call to the following Php variables
<?php
$scheduleStartDate = '';
$scheduleEndDate = '';
?>
Any help my greatly appreciated.
$("#schedule").change(function() {
var value = $('#schedule option:selected').text();
var ajaxCheck = $.ajax({
url: 'processClg.php',
type: 'POST', // had mention post bcoz u mention in processClg.php
dataType: 'json', // processClg.php will return string means change to text
data: { id: value },
success: function(data){
console.log('success');
itrToRead(data);
}
});
});
function itrToRead(data) {
$(data).each(function(key, value){
console.log('key is: '+key+' and value is: '+value);
});
}
processClg.php
<?php
include "config.php";
if ($_POST['type']=='POST') {
$qry = "SELECT * FROM schedule WHERE scheduleName LIKE '".$_POST['id']."'";
$res = mysqli_query($link,$qry);
echo $res;
}
?>
You can call ajax as follows:
var RowId;
$.ajax({
type: "POST",
async: false,
url: url,
data: postdata,
//dataType: "json",
success: function (data) {
RowId = data;
}
});
The fact is that to assign response to variable is pass the parameter async: false,
Then its work.
I guess you have your dropdown in your rendered page and you want to send the selected value to the php page:
$("#schedule").change(function(){
var val2pass = $(this).find(':selected').val(); // get the value
$.ajax({
url: 'processClg.php',
type:'post', // <-----you need to use post as you are using $_POST[]
data: { rowid : val2pass }, //<---pass the value
success: function(data){
itrToRead(data);
}
});
});
So now on the php side you need to do this:
<?php
include "config.php";
if ($_POST['type']=='POST')
{
$qry = "SELECT * FROM schedule WHERE scheduleName LIKE '".$_POST['rowid']."'";
$res = mysqli_query($link,$qry);
}
?>
Your procellClg.php will be
<?php
include "config.php";
if (isset($_REQUEST['qid']))
{
$qry = "SELECT * FROM schedule WHERE scheduleName LIKE '".$_REQUEST['qid']."'";
$result = mysqli_query($link,$qry);
echo "<select name='schedule' id='schedule'>";
echo "<option value=''>-- Select Schedule --</option>";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['scheduleName'] . "'>" . $row['scheduleName'] . "</option>";
}
echo "</select>";
}
?>
and then make Ajax function call
$("#schedule").change(function() {
val = $(this).val();
$.ajax({
type: "POST",
async: false,
url: 'processClg.php',
data: {qid:val},
success: function(data){
$(this).html(data);
}
});
});
I would check if my JSON response (data) is empty
$.ajax ({
type : 'POST',
url : get.php,
dataType: 'json',
success : function(data) {
console.log(data.lenght);
}
});
GET.PHP
$query = "SELECT * FROM TABLE";
$stmt = $this->db->mysqli->prepare($query);
$stmt->execute();
$result = $stmt->get_result();
$_assoc = array();
while($row = $result->fetch_array(MYSQLI_ASSOC)) {
$_assoc = $row;
}
$stmt->close();
echo json_encode($_assoc);
JSON RESPONSE
{"foo":"6","bar":"3436","id":4,"code":""}
EMPTY JSON RESPONSE
[]
I tried with data.length but returned undefined.
UPDATE
I solved this way.
Object.keys(data).length
Thanks for yours downvote!
I have jquery pop form . It takes one input from the user ,mapping_key , Once the user enters the mapping key ,i make an ajax call to check if there is a user in the database with such a key.
This is my call .
Javascript:
$.ajax({
url : base_url+'ns/config/functions.php',
type: 'POST',
data : {"mapping_key":mapping_key} ,
success: function(response) {
alert(response)
}
});
PHP:
$sql = "select first_name,last_name,user_email,company_name from registered_users where mapping_key = '$mapping_key'";
$res = mysql_query($sql);
$num_rows = mysql_num_rows($res);
if($num_rows == 0)
{
echo $num_rows;
}
else{
while($result = mysql_fetch_assoc($res))
{
print_r($result);
}
}
Now i want to loop through the returned array and add those returned values for displaying in another popup form.
Would appreciate any advice or help.
In your php, echo a json_encoded array:
$result = array();
while($row = mysql_fetch_assoc($res)) {
$result[] = $row;
}
echo json_encode($result);
In your javascript, set the $.ajax dataType property to 'json', then you will be able to loop the returned array:
$.ajax({
url : base_url+'ns/config/functions.php',
type: 'POST',
data : {"mapping_key":mapping_key} ,
dataType : 'json',
success: function(response) {
var i;
for (i in response) {
alert(response[i].yourcolumn);
}
}
});
change
data : {"mapping_key":mapping_key} ,
to
data: "mapping_key=" + mapping_key,
You have to take the posted mapping_key:
$mapping_key = $_POST['mapping_key'];
$sql = "select first_name,last_name,user_email,company_name from registered_users
where mapping_key = '$mapping_key'";
or this:
$sql = "select first_name,last_name,user_email,company_name from registered_users
where mapping_key = $_POST['mapping_key']";
I have this query
$result = mysql_query("SELECT * FROM ship_data WHERE id = $ship") or die(mysql_error());
$rows = array();
while($r = mysql_fetch_assoc($result)) {
$rows = $r;
echo json_encode($rows);
}
And this bit of ajax to return the results
$.ajax({
type: "POST",
dataType: "json",
data: "ship=" + ship,
cache: false,
url: "/getdata.php",
success: function (data) {
alert(data.carrier);
}
});
If there is just one result in the array it works, if the array has multiple results in the array nothing is alerted.
Change your while loop to...
$rows = array();
while($r = mysql_fetch_assoc($result)) {
$rows[] = $r;
}
echo json_encode($rows);
And your JavaScript should always iterate over data with $.each().
I have a two jquery function to bring ratings and comments separately from two files which is working fine.
Now i want to do it in a single ajax call, i am trying to merge two function together this way but its not working.
jquery
function get_review(){
$.ajax({
type: "POST",
url: '../review.php',
data: {value1:value1, value2:value2, value3:value3},
dataType: 'json',
cache: false,
success: function(data)
{
var x = data[0];
var rating = (x-0.5)*2
var y = (20 * rating)+40;
$('#urating').css("backgroundPosition","0%" +(y)+ "px");
$('#comments').html(data);
}
});
};
PHP
$find_data = "SELECT * FROM $tablename WHERE table_name='$table' AND product_id='$id' ORDER by id DESC";
$query = mysqli_query($connection, $find_data);
$find_data2 = "SELECT * FROM $tablename2 WHERE id='$id'";
$query2 = mysqli_query($connection, $find_data2);
$row2 = mysqli_fetch_assoc($query2);
header('Content-type: application/json');
echo json_encode($row2);
?>
<?php while($row = mysqli_fetch_assoc($query)):?>
<div class="comment-container">
<div class="user-info"><?php echo $row['user_name']; ?></div>
<div class="comment"><p><?php echo $row['quick_comment']; ?></p></div>
</div>
<?php endwhile;?>
Please see and suggest any possible way to do it.
Thanks.
Try to encode whole response to JSON object!
Something like:
$response = array(
'success' => true,
'object' => $yourobject_or_array,
'html' => '<b>Bla bla</b>'
);
echo json_encode($response);
die();
JS:
function(response) {
var res = false;
try {
res = jQuery.parseJSON(response);
} catch(e) {}
if (res && res.success) {
// Use res.object and res.html here
}
}