CODEWITHSUNDEEP
phpjavascriptjqueryajaxjson
Gásten
Gásten

Reputation: 133

User provided array name in javascript

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?

Upvotes: 0

Views: 84

Answers (3)

SeanCannon
SeanCannon

Reputation: 78006

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'.

Upvotes: 3

dfsq
dfsq

Reputation: 193311

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));

Upvotes: 0

Dr.Molle
Dr.Molle

Reputation: 117354

use:

var image_loc = data[nome][7];

Upvotes: 1

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