Transferring a multidimensional array from mysql to php to javascript

Question

I seem to be having a bit of trouble. The PHP portion seems to work well, but I can't seem to get the info over to the JS portion. Here is what I currently have:

<!DOCTYPE html>
<html> 
  <body>
    <?php
    $servername = "localhost";
    $username = "username";
    $password = "password";
    mysql_connect($servername, $username, $password) or die(mysql_error());
    mysql_select_db("stuff") or die(mysql_error());
    $result = mysql_query("SELECT Location, Latitude, Longitude FROM Places WHERE Latitude IS NOT NULL GROUP BY Location, Latitude, Longitude");
    $places = array();
    while (($row = mysql_fetch_array($result, MYSQL_NUM)) !== false) {
        $places[] = $row;
    }
    ?>
    <script>
    function makeTableHTML(myArray) {
        var result = "<table border=1>";
        for(var i=0; i<myArray.length; i++) {
            result += "<tr>";
            for(var j=0; j<myArray[i].length; j++){
                result += "<td>"+myArray[i][j]+"</td>";
            }
             result += "</tr>";
        }
    result += "</table>";
    return result;
    }
    var places = "<?php echo json_encode($places) ?>";
    makeTableHTML(places);
    </script>
  </body>
</html>

I have tested the PHP portion and it works just fine. I am fairly new to the JS and I am not sure how I can go about testing/debugging this. Any help would be appreciated.

I found the makeTableHTML function on here in hopes that it would help, but I didn't have any luck.

lal12 · Accepted Answer · 2016-02-18 01:19:27Z

1

Your PHP seems to be working. Also your javascrip HTML code generation seems to be ok. But you are no where writing the table into the DOM.

Therefore you have to select an DOM Element and write the result of makeTableHTML into it. For example:

document.getElementsByTagName("BODY")[0].innerHTML += makeTableHTML(places);

In your code example it would make more sense to generate the table with PHP directly. Javascript only make sense if you want to dynamically change your site.

I don't know what you are trying to archieve here, but maybe you should look into ajax requests. But it is hard to say, without knowing what you want to create.

You also asked about howto debug javascript, of course you could use alert() to output any variable in your code. But in more complex cases it is advisable to use the debugger. See here for howto use it in Firefox.

answered Feb 18, 2016 at 1:19

lal12

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8 Comments

rjbogz Over a year ago

Thanks for the info. I will be putting this into a Google Maps API via JS. In the end, I just need the info in an array to plot map points. It works if I write the array manually, but doesn't work if I use this. That led me to believe that array wasn't getting passed over. I am new to the whole DOM thing, so I will have to read up on it.

lal12 Over a year ago

Ok, then your approach seems to be good. As I already mentioned you also should have a look at the debugger. If you already have worked with a debugger you shouldn't have any trouble in using it. There you can also manipulating the DOM or running javascript commands at runtime.

rjbogz Over a year ago

I can't seem to get this to work, even with the debugger. The 2D array can't be seen by it. It just appears to be the PHP script I have it set to. Maybe I am missing something.

lal12 Over a year ago

Would you please post the output of the php script? Open the page in the browser and show the source code, then copy it into a jsfiddle e.g.

rjbogz Over a year ago

I am not sure how to go about doing that. I can do a print_r() in PHP, but I am trying to do a json_encode of that. I think the issue is with the transfer from PHP to JS. I googled around a bit, but everywhere seems to say that I am doing it correctly.

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