Dynamic PHP content inside ajax success function

Question

I am trying to create dynamic php content on ajax success like the one below. I keep getting parse error. Not able to figure out the issue. New to PHP n Ajax

$.ajax({
    type: "GET",
    url: "edit_listing.php",
    data: {value:val},
    dataType: "JSON",
    success: function(data) { 
        txt="<php echo '<div></div>';"+
            "echo '<p></p>'; ?>";
        $("#someid").append(txt);
    }
});

Mayank Pandeyz · Accepted Answer · 2017-02-03 08:39:55Z

1

Try this:

txt = '<div></div> <p></p>';
$("#someid").append(txt);

Explanation: You can create html and show it on page by using JS or JQuery only. There is no requirement of PHP in that.

answered Feb 3, 2017 at 8:39

Mayank Pandeyz

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5 Comments

Vinay Emmaadii Over a year ago

I want even to use some if conditions n more, this was just an example. I am building something which uses session variables for more dynamic content.

Mayank Pandeyz Over a year ago

Return the value of session variable from ajax call and use in JS

Vinay Emmaadii Over a year ago

I thinks thats better idea. Have done that before. Just wondering is there a way to use some thing like this code directly in ajax o/p "<?php if(isset($_SESSION['pictures']) && !empty($_SESSION['pictures'])) { echo '<div class="w3-content1" style="width:300px;height:250px;position:relative;overflow:hidden">'; for($i=0;$i<count($_SESSION['pictures']);$i++){ $value=$_SESSION['pictures'][$i]['picture_link']; if($i==0){ echo "<img src=$value class='mySlides' alt='Image' style='width:100%;height:100%;'>"; } }echo '</div>'; }?>

Tom Over a year ago

There's not. Stick to the way you've done it before.

Vinay Emmaadii Over a year ago

Thanks you all. If thats best way. I will stick to that.

ramabarca · Accepted Answer · 2017-02-03 15:09:35Z

0

As an alternative you can do it with json_encode. For example:

// php code
<?php $txt = "<div>{$somevar}</div><p>{$anothervar}</p>"; ?>

$.ajax({
    type: "GET",
    url: "edit_listing.php",
    data: {value:val},
    dataType: "JSON",
    success: function(data) {
        // php code 
        txt=<?php echo json_encode($txt); >;
        $("#someid").append(txt);
    }
});

answered Feb 3, 2017 at 15:09

ramabarca

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