CODEWITHSUNDEEP
phpjqueryjsonajax
user6930268
user6930268

Reputation:

How to display multiple values from AJAX in PHP?

I have to display multiple values from AJAX but I am getting only single value.

Concept: The user will select single or multiple check box and according to the checkbox, it will display the image and user_id. I am able to display the image but not able to display the user_id. Would you help me in this?

I just upload only AJAX and PHP code

AJAX

$.ajax({
    type: "POST",
    url: "includes/compare_process.php", // 
    data:'users='+arr,
    dataType: 'json',
    success: function(msg){
        $("#pics_name").empty();
        $.each(msg, function() {
            $("#pics_name").append("<img src='images/profile/" + this + "' alt='' />");
            $("#pics_Id").append();//I have to display id here
        });
    },
    error: function(){
        alert("failure");
    }
});

PHP

$_SESSION['compare_user']=$_POST['users'];
$sql_compare='SELECT * FROM request WHERE Id IN (' .( is_array( $_SESSION['compare_user'] ) ? implode( ',', $_SESSION['compare_user'] ) : $_SESSION['compare_user'] ).')';
$compare_query=$conn->query($sql_compare);
if ($compare_query->num_rows > 0) {
    while($userdata12=$compare_query->fetch_assoc()){ 
        $compare_pic[]=$userdata12['profile_pic'];
        $compare_Id=$userdata12['Id'];
    }
}
echo json_encode($compare_pic, $compare_Id);
exit();

Upvotes: 0

Views: 912

Answers (3)

Bilal Akbar
Bilal Akbar

Reputation: 4930

$.ajax({
    type: "POST",
    url: "includes/compare_process.php", // 
    data:'users='+arr,
    dataType: 'json',
    success: function(msg){
        $("#pics_name").empty();
        $.each(msg, function (index, value) {
            $("#pics_name").html("<img src='images/profile/" + value.profile_pic + "' alt='' />");
            $("#pics_Id").html(value.Id);
        });
    },
    error: function(){
        alert("failure");
    }
});

PHP 

$_SESSION['compare_user']=$_POST['users'];
$sql_compare='SELECT * FROM request WHERE Id IN (' .( is_array( $_SESSION['compare_user'] ) ? implode( ',', $_SESSION['compare_user'] ) : $_SESSION['compare_user'] ).')';
$compare_pic = array();
$compare_query=$conn->query($sql_compare);
if ($compare_query->num_rows > 0) {
     while($userdata12=$compare_query->fetch_assoc()){ 
            $compare_pic[]=$userdata12;
     }
   }
echo json_encode($compare_pic);
exit();

This should get you started. The above code is returning single array as JSON. In JavaScript, you can display any required property where needed.

Upvotes: 0

Barmar
Barmar

Reputation: 781028

You're not putting $compare_Id into an array. You need to create a 2-dimensional array in the PHP.

$_SESSION['compare_user']=$_POST['users'];
$sql_compare='SELECT id, profile_pic FROM request WHERE Id IN (' .( is_array( $_SESSION['compare_user'] ) ? implode( ',', $_SESSION['compare_user'] ) : $_SESSION['compare_user'] ).')';
$compare_pic = array();
$compare_query=$conn->query($sql_compare);
if ($compare_query->num_rows > 0) {
     while($userdata12=$compare_query->fetch_assoc()){ 
            $compare_pic[]=$userdata12;
     }
   }
echo json_encode($compare_pic);
exit();

Then access each property in the Javascript.

$.ajax({
    type: "POST",
    url: "includes/compare_process.php", // 
    data:'users='+arr,
    dataType: 'json',
    success: function(msg){
        $("#pics_name,#pics_Id").empty();
        $.each(msg, function() {
            $("#pics_name").append("<img src='images/profile/" + this.profile_pic + "' alt='' />");
            $("#pics_Id").append("<div>" + this.id + "</div>");
        });
    },
    error: function(){
        alert("failure");
    }
});

Upvotes: 0

Mahdi P.
Mahdi P.

Reputation: 307

the second argument in function jsonencode is for options according following link:

http://php.net/manual/en/function.json-encode.php

for passing multiple values you can put your variables into an array then put this array in function jsonencode

so you can do something like this in your php code:

$myArray = [
     'compare_pic' => $compare_pic,
     'compare_Id'  => $compare_Id
];

echo json_encode($myArray);

then you get each value in your response in jQuery code like this:

 var firstValue  = msg.compare_pic
 var secondValue = msg.compare_Id

Upvotes: 1

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