Populating a JavaScript array with MySql data

Question

I've looked through various other questions on SO, but can't quite get the right answer. Basically, I have an array that needs data from a MySql database. Here's what I've tried:

var $name = "Foo",
    $x = 10,
    $bar =
      <? php
        $barQuery = mysql_query("SELECT item FROM table WHERE name = '$name' AND number = '$x'");
        $barArray = array();

        while ($r = mysql_fetch_assoc($barQuery))
        {
          $barArray[] = $r['item'];
        }

        echo json_encode($barArray);
      ?>;

EDIT: I then post this to a .php file, and print the returned data:

$.post("file.php", {bar: JSON.stringify($bar), name: $name}).done(function(data)
{
  $('body').html(data);
  window.print();
  setTimeout("location.reload(true)", 500);
});

However, I get an error saying "syntax error, unexpected T_VARIABLE". Is it not possible to populate a JS array this way, or is there another way to do it?

@AdrianCidAlmaguer The reason I use the dollar sign, is because I then post this information to another .php file using `$.post();` — Michael Jarvis, Mar 07 '15 at 13:58
PHP syntax errors usually specify the line number on which the error occurred—this is extremely helpful information (although you will have to map that to your code snippet for us). — eggyal, Mar 07 '15 at 14:00
If `$name` and `$x` are JavaScript variables, then you cannot access them within PHP (which is processed on the server-side, before the JavaScript is even transmitted to the client): therefore they are not defined when used in your SQL. — eggyal, Mar 07 '15 at 14:03
@eggyal I could use `$.post()` to post the variables, but how would I then put the returned data into a JS array variable? The same way I do here using `echo`? — Michael Jarvis, Mar 07 '15 at 14:15
@eggyal The error is showing up at the first line of php, so the line that has ` php` — Michael Jarvis, Mar 07 '15 at 14:21
@AdrianCidAlmaguer Yes, but it's only an `if` statement, which this code is nested inside — Michael Jarvis, Mar 07 '15 at 14:35
@AdrianCidAlmaguer Yes, and am getting the same error as commented on eggyal's answer — Michael Jarvis, Mar 07 '15 at 14:40
Yes, I connect to the database before the HTML head, and I also interact with the database in the HTML which works fine. The only issue is this part in the JS. Would it be easier (and better) to use `jQuery.post()` and then use the return data, or stick with it this way? — Michael Jarvis, Mar 07 '15 at 14:47
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/72472/discussion-between-adrian-cid-almaguer-and-michael-jarvis). — Adrian Cid Almaguer, Mar 07 '15 at 14:48

Adrian Cid Almaguer · Answer 1 · 2015-03-07T14:00:43.333

0

var $name = "Foo",
    $x = 10,
    $bar = "
      <?php
        $barQuery = mysql_query("SELECT item FROM table WHERE name = '$name' AND number = '$x'");
        $barArray = array();

        while ($r = mysql_fetch_assoc($barQuery))
        {
          $barArray[] = $r['item'];
        }

        echo json_encode($barArray);
      ?>";

edited Mar 07 '15 at 14:00

answered Mar 07 '15 at 13:54

Adrian Cid Almaguer

7,815
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Would this cause any issues since the speech marks cancel out where the SELECT begins? – Michael Jarvis Mar 07 '15 at 14:13
I have edited the OP to show what happens next. What else is required? – Michael Jarvis Mar 07 '15 at 14:24
@MichaelJarvis you got some advance? – Adrian Cid Almaguer Mar 08 '15 at 03:48

score 0 · Answer 2 · answered Mar 07 '15 at 14:24

0

You have a superfluous space inside the PHP opening tag:

      <? php
        ^--- this shouldn't be there

As it stands, the <? is parsed as a short open tag, and so the php is parsed as a(n undefined) constant, which is immediately followed by $barQuery—hence the syntax error that you see: unexpected T_VARIABLE.

answered Mar 07 '15 at 14:24

eggyal

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It's always something so simple... Thanks! However, an error in the JS console is saying there is an 'unterminated string literal' when I declare `$bar` – Michael Jarvis Mar 07 '15 at 14:27
@MichaelJarvis: That's strange. Have you changed your code from what it originally was in your question? Check the JavaScript that is output by PHP (and therefore parsed by the browser); your JS console might help to pinpoint the problem. – eggyal Mar 07 '15 at 14:29
Yes, here's what it said: `$bar = "
Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in /home/page.php on line 791
[]";` Line 791 is where the `while` loop begins – Michael Jarvis Mar 07 '15 at 14:34
@MichaelJarvis: That means that `$barQuery` is false, because `mysql_query()` failed. You should inspect `mysql_error()`. See http://stackoverflow.com/questions/2973202/mysql-fetch-array-expects-parameter-1-to-be-resource-or-mysqli-result-boole – eggyal Mar 07 '15 at 14:49

Populating a JavaScript array with MySql data

2 Answers2