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POwrong output in address array after using json_encode() and JSON.parse
 

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  1. POwrong output in address array after using json_encode() and JSON.parse
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    <p>I have this PHP which is fetching all rows of address column from the database and i an using <code>json_encode()</code> function to convert it into string and <code>JSON.parse</code> in javascript but i am not getting output as expected</p> <pre><code> &lt;? include('dbcon.php'); $result = mysql_query("SELECT address FROM markers"); while ($row = mysql_fetch_assoc($result)) { $new_array[] = $row; } print_r($new_array); $add_js = json_encode( $new_array ); print_r($add_js); ?&gt; </code></pre> <p>with <code>print_r($new_array);</code> i am getting two dimentional arrary and getting <code>null</code> in zeroth location. any changes i make it keeps on giving null at zeroth location. </p> <pre><code> var address = [JSON.parse( '&lt;?php echo $add_js ?&gt;' )]; </code></pre> <p><code>var address</code> should store array of addresses but it is giving output when i alert in the array.</p> <p><code>[object Object],[object Object],[object Object],[object Object],[object Object]</code> </p> <p>i want to store array of addresses after fetching them from database </p>
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    1. USanalyticalpicasso
    1. USanalyticalpicasso
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      1. POwrong output in address array after using json_encode() and JSON.parse
      1. USanalyticalpicasso
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    2. VO
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      1. VTUpMod
    1. COCan you please post the results which your PHP generates?
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      1. USBergi
    2. COFirst of all initialize the $new_array by adding $new_array = arary() at the begining of you code. Then alert won't display json string, it is an object, rather than using alert() for debuging, add console.log(address); and checkout your browser console (f11).
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      1. USFilip Górny
    3. COWhat's wrong with that array of objects, what did you expect (`[object Object]` is OK for an object that gets stringified from an alert)? Are you sure that you need the brackets (array literal) around the `JSON.parse` expression? Also, you should not need to use a string and explicit parse, you can simply echo the JSON as a javascript literal into the code.
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      1. USBergi
 

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