User provided array name in javascript

Question

So i am fetching rows from my database using AJAX and then turning them into an array with a variable identifier Here is the code PHP:

 $query_val = $_GET["val"];
  $result = mysql_query("SELECT * FROM eventos_main WHERE nome_evento LIKE '%$query_val%' OR local_evento LIKE '%$query_val%' OR descricao_evento LIKE '%$query_val%'");
  for($i=0;$i<mysql_num_rows($result);$i++){
    $array = array();
    $array[$query_val] = mysql_fetch_row($result);       //fetch result 
    echo json_encode($array);
  }

Here is the javascript:

$('#s_query').keyup(function(){
        var nome = document.getElementById('s_query').value;
        $.ajax({                                      
            url: 'search.php',        
            data: '&val='+nome,
            dataType: 'json',     
            success: function(data)
        {
            console.log(nome);
            var image_loc = data.nome[7];
            console.log(image_loc);

If i change the line var image_loc = data.nome[7]; to var image_loc = data.nidst[7]; it works perfectly. Nidst is the term i search for. This code returns the error: "Uncaught TypeError: Cannot read property '7' of undefined". What should i do?

Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — Joseph Silber
– Joseph Silber, Commented Jan 21, 2013 at 19:18
Yes i saw it on the php website, but i'll update my code later when i have everything done. — Gásten
– Gásten, Commented Jan 21, 2013 at 19:58

SeanCannon · Accepted Answer · 2013-01-21 19:34:44Z

3

data.nome[7]; is trying to access a property of data named nome, which doesn't exist. Since you declared a variable nome which holds your desired property name, you need to reference the value as the property name, like data[nome][7].

Example: If var nome = 'foo', then data[nome][7] will evaluate to data['foo'][7] which is the same as data.foo[7].

What you are doing is data.nome[7] which is the same as data['nome'][7], and the only way that would work is if var nome = 'nome'.

edited Jan 21, 2013 at 19:34

answered Jan 21, 2013 at 19:28

SeanCannon

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Dr.Molle · Accepted Answer · 2013-01-21 19:21:59Z

1

use:

var image_loc = data[nome][7];

answered Jan 21, 2013 at 19:21

Dr.Molle

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5 Comments

user400654 Over a year ago

Why is this downvoted? it's the correct answer.(for the javascript side of the problem atleast, this doesn't address the invalid json being returned when there are more than 1 records)

adeneo Over a year ago

This is obviously one of the problems, not sure why it was downvoted? +1 .

Ian Over a year ago

I didn't downvote, but wouldn't it actually be helpful if this was explained? Some people, unfortunately, may not understand why data.nome wouldn't work compared to data[nome].

Dr.Molle Over a year ago

It's true, I didn't address the possibly invalid JSON, because this wasn't the issue here(data.nidst[7] returns a result, so data must be valid JSON) . But of course the JSON-issue has to be fixed too.

SeanCannon Over a year ago

I'm not seeing a downvote here. 1 upvote, 0 downvotes. I haven't even started drinking yet... lol

dfsq · Accepted Answer · 2013-01-21 19:28:08Z

0

Another problem is that server does not resond with valid json data. Try to modify the code:

$array = array();
for ($i = 0; $i < mysql_num_rows($result); $i++) {
    $array[$query_val] = mysql_fetch_row($result);
}
die(json_encode($array));

answered Jan 21, 2013 at 19:28

dfsq

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