Error in update and insert query in php mysql

Question

I am trying to update seen(a bool in the database) if the seen button is clicked and delete a row if the delete button is clicked. But it doesn't seem to work nor do I get any data(mysql_error) back. But the last else part works and I get the data.

This is the php code :

if(isset($_POST["seen"])){
    $fid=$_POST["id"];
    $sql="UPDATE feedback SET seen=1 WHERE ID=$fid";
    mysql_query($sql,$con);
    echo mysql_error($con);
} else if(isset($_POST["delete"])){
    $fid=$_POST["id"];
    $sql="DELETE from feedback WHERE ID=$fid";
    mysql_query($sql,$con);
    echo mysql_error($con);
} else{
    $sql="SELECT * FROM feedback";
    $result=mysql_query($sql,$con);
    while($row=mysql_fetch_array($result)){
        $jsonData[]=$row;
    }
    echo json_encode($jsonData);
}

and js code:

$("#seen").click(function(){
    $.post("main_php.php",{},function(data){
        alert(data);
    }); 
});

$("#delete").click(function(){
    var fid = $('#ID').val();
    $.post("main_php.php",{id:fid},function(data){
        console.log(data);
    }) 
});

Please tell me what may be wrong with it!!!!

seen and delete are the names of buttons btw

In the name of all that is good and beautiful in this world, just stop writing new PHP code that uses the deprecated mysql_ interface functions. It's 2016 already. Use PDO or mysqli. And prepared statements with bind placeholders are not that hard. Really. And I'm so going to pass in the value '1 OR 1=1 --' for id. — spencer7593
– spencer7593, Commented Jan 16, 2016 at 4:40

Jason Kennaly · Accepted Answer · 2016-01-16 04:05:06Z

2

Your jQuery POST is not setting the "seen" or "delete". So when the PHP executes, it is just falling through to that last else.

In the jQuery, set "seen" or "delete" to true (or anything else, but they need to be set) and you should be good.

answered Jan 16, 2016 at 4:05

Jason Kennaly

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1 Comment

user1881473 Over a year ago

seen and delete are buttons.Are they not set when clicked upon?

nikamanish · Accepted Answer · 2016-01-16 04:10:28Z

0

Set the 'seen' / 'delete' in jQuery to true as they are not set. That might solve the problem.

answered Jan 16, 2016 at 4:10

nikamanish

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1 Comment

user1881473 Over a year ago

seen and delete are buttons.Are they not set when clicked upon?

Thanga · Accepted Answer · 2016-01-18 06:27:05Z

0

In Ajax submit, we have to explicitly submit the elements you want. To fix it, edit your code like below

$("#seen").click(function(){
    $.post("main_php.php",{seen:1},function(data){
        alert(data);
    }); 
});

$("#delete").click(function(){
    var fid = $('#ID').val();
    $.post("main_php.php",{id:fid,delete:1},function(data){
        console.log(data);
    }) 
});

edited Jan 18, 2016 at 6:27

answered Jan 16, 2016 at 4:10

Thanga

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5 Comments

user1881473 Over a year ago

seen and delete are buttons.Are they not set when clicked upon?

Thanga Over a year ago

Those buttons will be submitted when the form submission happens. But what you are doing is a Ajax submit in which we have explicitly submit the elements you want

user1881473 Over a year ago

can you help me with one more question pls

user1881473 Over a year ago

just search 'Display dynamic data using jquery chart' and check the first post

user1881473 Over a year ago

stackoverflow.com/questions/34823961/…

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