json_encode in php

Question

i have this script and it works, but not as i expected. I need to assign an array of values with differents names, now all $arr[] are named "valor"

{"valor":"20"},{"valor":"50"}

i need

{"valor1":"20"},{"valor2":"50"}

the script

    $query = mysql_query("SELECT valor FROM grafico") or die(mysql_error());

    $arr = array();
    while($row = mysql_fetch_assoc($query)) {
        $arr[] = $row;
    }
    echo json_encode($arr);


in ajax

    <script type="text/javascript">
       jQuery(document).ready(function(){
          jQuery("button").click(function(){

             jQuery.ajax({
                url: "chart.php",
                dataType: "json", 
                success: function(json){
                   var msg = "Nome: " + json.valor1+ "\n";
                   msg += "Sobrenome: " + json.valor2 + "\n";

                   alert(msg);
                }
             });
          });
       });

       </script>

the problem is: I need to create a loop that create uniques names, as value1, value2, value3

like

$arr['valor1'] = "Evandro";

i tried a loop- for, but i get an error of memory

thanks

Michał Niedźwiedzki · Accepted Answer · 2011-03-24 10:35:12Z

2

Try this:

$query = mysql_query("SELECT valor FROM grafico") or die(mysql_error());

$arr = array();
$i = 1;
while($row = mysql_fetch_assoc($query)) {
    $arr[] = array("valor{$i}" => $row["valor"]);
    ++$i;
}
echo json_encode($arr);

Should work. Alternatively if you want to make it so it works with current callback change the $arr[] = line to the following:

    $arr["valor{$i}"] = $row["valor"];

edited Mar 24, 2011 at 10:35

answered Mar 23, 2011 at 23:27

Michał Niedźwiedzki

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5 Comments

daniel__ Over a year ago

the result is [{"valor1":"20"},{"valor2":"50"}] ?> but the alert doesn't work. It is supposed work, right?

Michał Niedźwiedzki Over a year ago

You sure it ought to be jQuery("button") and not jQuery("#button")? Also, try alert()-ing json var at the very top of callback function.

daniel__ Over a year ago

the alert works but says valor1: undefined and the same to the valor2. Thanks

daniel__ Over a year ago

the problem is the [] --- [{"valor1":"20"},{"valor2":"50"}], but if i remove in the [] in $arr only show the second value {"valor2":"50"}

Michał Niedźwiedzki Over a year ago

Updated the answer to work with your callback. Just a side note - don't call your variable "valor" ("value" in Spanish, I'm guessing) - it could mean anything. It's like calling your dog "Dog". It's plain to see this is some kind of value. The question is what kind exactly? If it's name, call it "name". If it's user details call it accordingly, etc.

Chris · Accepted Answer · 2011-03-23 23:27:08Z

1

$index = 1;
while($row = mysql_fetch_assoc($query)) {
        $key = 'valor'.index;
        $arr[$key] = $row;
        $index++;
    }

Does that give you a memory error? It shouldn't.

answered Mar 23, 2011 at 23:27

Chris

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1 Comment

Chris Over a year ago

Unless you have an unbelievably immense number of return values from your query.

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