No Ajax success after php-script is run

Question

I have a php-script which performs a database backup of my website. This backup-script can be triggered by clicking a button on the websites' admin-panel.

I'd like to have some text in the admin-panel that tells the user when the backup is ongoing and when it's complete (process takes about 1 minute) in real-time.

I believe using Ajax is the only reasonable way of doing this, and I have tried the following so far:

<button id="btn">Backup</button>
<script>
    $("document").ready(function(){
        $("#btn").click(function(){
            $.ajax({
                type: "POST",
                dataType: "json",
                url: "backup/db_backup.php",
                success: function(data) {
                    alert("success!!!");
                }
            });
        });
    });
</script>

The above does run the backup-script just fine, but I'm unsure why the alert doesn't trigger once the script has finished running. Is there something specific in the php-file I need to have for this to work?

Any help is much appreciated!

EDIT: added the php script as requested

<?php
    include_once '../includes/db_connect.php';

    ini_set("max_execution_time", 0);

    $zip = new ZipArchive();
    $dir = "backups";

    if(!(file_exists($dir))) {
        mkdir($dir, 0777);
    }

    $p = backup_tables($mysqli);
    echo $p;
    if (glob("*.sql") != false) {
        $filecount = count(glob("*.sql"));
        $arr_file = glob("*.sql");

        for($j=0;$j<$filecount;$j++) {
            $res = $zip->open($arr_file[$j].".zip", ZipArchive::CREATE);
            if ($res === TRUE) {
                $zip->addFile($arr_file[$j]);
                $zip->close();
                unlink($arr_file[$j]);
            }
        }
    }

    //get the current folder name-start
    $path = dirname($_SERVER['PHP_SELF']);
    $position = strrpos($path,'/') + 1;
    $folder_name = substr($path,$position);
    //get the current folder name-end

    $zipname = date('Y/m/d');
    $str = "dRbot-".$zipname.".zip";
    $str = str_replace("/", "-", $str);

    // open archive
    if ($zip->open($str, ZIPARCHIVE::CREATE) !== TRUE) {
        die ("Could not open archive");
    }

    $zip->addFile(realpath($folder_name . "/" . $p));
    // close and save archive
    $zip->close();
    echo "Archive created successfully.";

    copy("$p.zip", "$dir/$str");
    unlink("$p.zip");

    /* backup the db OR just a table */
    function backup_tables($mysqli, $tables = '*') {
        //get all of the tables
        if($tables == '*') {
            $tables = array();
            $result = $mysqli->query('SHOW TABLES');
            while($row = $result->fetch_array(MYSQLI_NUM)) {
                $tables[] = $row[0];
            }
        } else {
            $tables = is_array($tables) ? $tables : explode(',',$tables);
        }
        $return = "";

        //cycle through
        foreach($tables as $table) {
            $result = $mysqli->query('SELECT * FROM '.$table);
            $num_fields = mysqli_field_count($mysqli);
            $return.= 'DROP TABLE '.$table.';';
            $result2 = $mysqli->query('SHOW CREATE TABLE '.$table);
            $row2 = $result2->fetch_row();
            $return.= "nn".$row2[1].";nn";

            while($row = mysqli_fetch_row($result)) {
                $return.= 'INSERT INTO '.$table.' VALUES(';
                for($j=0; $j<$num_fields; $j++) {
                    $row[$j] = addslashes($row[$j]);
                    $row[$j] = preg_replace("#n#","n",$row[$j]);
                    if (isset($row[$j])) { $return.= '"'.$row[$j].'"' ; } else { $return.= '""'; }
                    if ($j<($num_fields-1)) { $return.= ','; }
                }
                $return.= ");n";
            }
            $return.="nnn";
        }

        //save file
        $path = 'db-backup-'.time().'-'.(md5(implode(',',$tables))).'.sql';
        $handle = fopen($path,'w+');
        fwrite($handle,$return);
        fclose($handle);
        return $path;
    }

?>

@AhosanKarimAsik, I can provide this if necessary, but I believe there isn't anything to see there that's relevant to the Ajax post. The script creates a `.sql` file and then zips it -- nothing else. — Chris, Jun 14 '15 at 20:22
try with alert(data). If there is any error server side you will probably see it in the alert box. Event better go for console.log(data) — Fabrizio Mazzoni, Jun 14 '15 at 20:24
In your php code just after doing all stuff put `echo 'success';` and then run this script and check what happen. thanks — Alive to die - Anant, Jun 14 '15 at 20:25
@FabrizioMazzoni, I tried this but I get no alert at all. I'll try console log... — Chris, Jun 14 '15 at 20:29
Wont work because data is the same.. Are you at least echoing something when the php script has finished? Would be interesting to see the script itself. If the script completes successfully and there is nothing sent back, data will be empty, thus producing empty alert box.. — Fabrizio Mazzoni, Jun 14 '15 at 20:30
instead of `return $path;` try to do `echo $path;` and check. — Alive to die - Anant, Jun 14 '15 at 20:44
Maybe the browser is timing out? What does it say in the Network tab of the console? — Barmar, Jun 14 '15 at 20:47

score 1 · Accepted Answer · answered Jun 14 '15 at 20:35

1

Your script db_backup.php should output something in JSON format, echo something like this.

echo json_encode(array('success'));

OR

You can change your code to this:

<button id="btn">Backup</button>
<script>
    $("document").ready(function(){
        $("#btn").click(function(){
            $.ajax({
                type: "POST",
                url: "backup/db_backup.php",
                success: function(data) {
                    alert("success!!!");
                }
            });
        });
    });
</script>

answered Jun 14 '15 at 20:35

antihate

479
6
16

I tried removing the json datatype as you suggested but still the same. – Chris Jun 14 '15 at 20:40
I created a jsfiddle, showing that code is working, here's the link https://jsfiddle.net/jL1knr66/ with both solutions I gave. – antihate Jun 14 '15 at 20:52
It works now, thanks. I did try your code before though and it didn't work -- perhaps the browser didn't fully refresh. – Chris Jun 14 '15 at 21:00

dgo · Answer 2 · 2015-06-14T23:42:42.360

I would try something like this:

 var serverCall=function(){
      var ajax=$.ajax({
            type: "POST",
            dataType: "json",
            url: "backup/db_backup"
        });


     return ajax.then(function(data,status,xhr){
                              // success
                }, function(){
                    // fail
                });

  }



 $("#btn").on('click',serverCall);

That should give you what you want. One thing to keep in mind is what you are returning from your server code. If you are returning false or true, it will have no impact on which callback handler is called. You need to send an HTTP status code in your response headers. That is the only way jQuery can determine if the request was successful.

Lastly. You only have a success callback. Your code should still work unless the server side returned a 4xx or 5xx response code. Then it will fire the fail callback handler, which you don't have, so nothing will happen. You can attach a function instead of then called always as a callback handler. This will fire a callback regardless of the outcome.

I recommend investigating what I said in the jQuery docs.

The success callback is not deprecated. – Barmar Jun 14 '15 at 20:44 — Barmar, Jun 14 '15 at 20:44

No Ajax success after php-script is run

2 Answers2