phpajax, SyntaxError: Unexpected identifier

Asked 2 years ago, Updated 2 years ago, 115 views

What I want to do is: I want to run the include/sync.php file when I press the send button.

Description:
I retrieve the data from mysql database in php and display it in the order of the latest id. I put buttons on each id, send the data, and run the php file.
What is wrong with the error below?
There are from id(1) to id(10), but when I press the latest id(10), the error below appears, but when I press the other id button, it doesn't respond at all!!
Why is that?Someone please teach me.

Description Supplement: /* The following question is an error when rewriting this with aax
This works fine, but =>echo "send";*/

Error below:

SyntaxError: Unexpected identifier

1, The above error will appear on the send button below

 echo "<td><arel='$test_id' href='javascript:void(0)'id='sw'>send</a></td>";

<script type="text/javascript" src="http://ajax.googleapi.com/ajax/libs/jquery/1.7.2/jquery.min.js">/script>
<script type="text/javascript">
    $(document).ready(function(){
        $("#sw").click(function(){
            varparam={"test_id":"{$test_id}"};
                $.ajax({
                type: "post",
                url: "includes/sync.php",
                data —JSON.stringify (param),
                dataType: "text",
            }).done(function(data){
                alert(data.text);
            }).fail(function(XMLHttpRequest, textStatus, errorThrown){
                alert(errorThrown);
            });
        });
    });
</script>

php jquery ajax

2022-09-30 20:12

1 Answers

The jQuery link seems to be wrong. Try fixing it to http://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js. (googleapi->googleapis)


2022-09-30 20:12

If you have any answers or tips


© 2024 OneMinuteCode. All rights reserved.