PHP json_encode return unescaped '

Question

I experience the problem with parsing json encoded object in js. JSON.parse(word_array); with error Uncaught SyntaxError: Unexpected identifier

My investigation showed that object word_array does not exist because of wrong-formation in PHP: it has a unescaped single quote' inside, thus making js consider it as the end of the string.

I form encoding next way:

echo "<script>var word_array = '";
echo  json_encode($word_set);
echo "';\n";

As far as I know, json_encode should escape all undesirable charactets like ' but it does not. What might be the problem?

My php version: Version PHP: 5.3.13 And $word_set is array of assoc. array:

$word_set = array();
while($stmt->fetch())
{
  $word_set_tmp[] = array(
    'word' => $word, 
    'definition' => $def
  );
  array_push ($word_set,$word_set_tmp);
}

Please show the output of json_encode, showing the claimed unescaped quote — Steve
– Steve, Commented Jul 29, 2014 at 14:26
Why would you assign string and then parse it when you could assign json right away (without wrapping it in quotes). — Cthulhu
– Cthulhu, Commented Jul 29, 2014 at 14:27
I cut the end (as it is big). var word_array = '[{"word":"Abeyance","definition":"a temporary suspension of activity."},{"word":"Akimbo","definition":"having one's hands in a bent position on the hips."} — user3365404
– user3365404, Commented Jul 29, 2014 at 14:27
@Cthulhu funny thing: your suggestion fixes his issue too, as afaik ' shouldnt be escaped — Valerij
– Valerij, Commented Jul 29, 2014 at 14:30

Steve · Accepted Answer · 2014-07-29 14:37:45Z

5

The problem is you surrounding an array declaration in single quotes, remove them and all is well:

echo "<script>var word_array = " . json_encode($word_set) . ";";

As a side note, i find when i must mix php with anything else (html, js) then its best to exit php mode and write html/js, echoing out the required php vars, rather than echo out html/js:

<?php 
    $word_set = $db->somfunc();
?>
<script>
    var word_array="<?php echo json_encode($word_set);?>";
    alert(word_array[1].definition);
</script>

edited Jul 29, 2014 at 14:37

answered Jul 29, 2014 at 14:31

Steve

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