CODEWITHSUNDEEP
phpjavascriptjqueryjson
user2765711
user2765711

Reputation: 33

json_encode() works for one string but not another?

EDIT:

Solved. Somebody commented with an answer, but then deleted it. Embarrassingly, the problem was fixed by adding semi-colons after the two json_encode lines. I still have no idea why string1 worked without a closing semi-colon and string2 did not, but it works, so… success?

/EDIT

Hi… fairly nooby programmer here.

Within PHP, I'm grabbing two arrays from my database. This works fine. I then implode each array into a string. This too works fine.

Later in the same file, in the javascript section of my code (featuring jQuery), I do:

string1 = <?php echo json_encode($string_1);?>
string2 = <?php echo json_encode($string_2);?>

Encoding string1 works perfectly… so long as I comment out the string2 line entirely. It kills my page.

I've tried ensuring that my string is utf8-encoded. I've tried ensuring that there are no backslashes or other problematic characters. Currently, I've got some dummy code that's designed to be as unproblematic as possible… still doesn't work.

Here's the skeleton version of it.

PHP:

$array1 = array();
$array2 = array();


$i = 1;
while($row = mysqli_fetch_assoc($query_result)) {
   $array1[$i] = $row['column1'];
   $array2[$i] = $row['column2'];
   $i++;
}

$string1 = implode(",", $array1);
$string2 = "oh_come_on_why_doesnt_this_work"; // implode(",", $array2);

And JS/HTML/JQuery:

var string1;
var string2;

$.getScript("js/set_functions.js", function(){ loadedScript() });

function loadedScript() {

    string1 = <?php echo json_encode($string1);?>
    string2 = <?php echo json_encode($string2);?>
    // string2 = <?php echo "'" . $string2 . "'";?> doesn't work either

    alert(string1); // works if I comment out both string2 = … lines
    alert(string2); // displays "undefined" if the first string2 = … line is commented…
    // …out, doesn't display at all if it isn't

}

STUCK PLEASE HELP.

Let me know if you need to see more of my code, or if there's anything else I could do to help make this question better. Thanks.

PS: I'm not a good programmer, I know. Please don't correct my other bad coding habits IN LIEU OF helping solve this problem. Thanks!

EDIT: Sorry, in response to Joe Fraumbach, that was an error in my copy-pasting. I had actually declared $array2 = array();

Upvotes: 1

Views: 3615

Answers (2)

LexLythius
LexLythius

Reputation: 1944

EDIT: You are missing semicolons at the end of the JS assignment expressions, which in some cases will cause parsing problems: both lines may become part of a single, multiassignment expression. Try adding them and see if that fixes your problem.

string1 = <?php echo json_encode($string1); ?>;
string2 = <?php echo json_encode($string2); ?>;

Note that json_encode only works with UTF-8. So if you're not sure that its input is UTF-8, utf8_encode() the strings you feed it, otherwise the whole string is rendered as null in PHP (and translated into an empty string upon concatenation).

Upvotes: 2

000
000

Reputation: 27247

$array2 is never defined. You defined $array1 = array();. Do the same for $array2.

Upvotes: 0

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