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up vote 3 down vote favorite

This code, which I've found somewhere online, works fine and it's kind of what I'm trying to accomplish:

$(document).ready(function(){
    var arrData = [ "j", "Q", "u", "e","r","y" ];   
    alert(jQuery.inArray("Q", arrData));
});

However, I have an array from a loop with php/mysql that I output and save like this:

$query = mysql_query("SELECT * FROM geo_orter");
            while(($row = mysql_fetch_assoc($query))){
                $i = $row['ort_id'];
                $result[$i] = $row['ortnamn'];
            };
            $allaOrterjson=json_encode($result);

Then I go ahead and do this, which works:

var allaOrter=<?php echo $allaOrterjson ?>;

    document.write(allaOrter[0] + allaOrter[1] + allaOrter[2]);

and gives me

undefinedAborremålaAbbjörnahall

Here's the issue:

I tried

$(document).ready(function(){
    alert(jQuery.inArray("Aborremåla", allaOrter));
});

but it results in "-1" (Not found).

I'm trying to find out the index nr of an item in the array. Any ideas?

Log result: loggen

share|improve this question
4  
Please, don't use mysql_* functions to write new code. They are no longer maintained and the community has begun deprecation process. See the red box? Instead you should learn about prepared statements and use either PDO or MySQLi. If you can't decide which, this article will help you. If you pick PDO, here is good tutorial. – orourkek Aug 10 '12 at 16:39
    
ohh, I had no idea mysql_ is moving out. – Alisso Aug 10 '12 at 16:44
    
allaOrter[0] is undefined - why is that so? Do a console.log(allaOrter) in your JavaScript and check the developer console in Chrome/Firefox etc. What does it say? – Anders Holmström Aug 10 '12 at 16:44
    
@AndersHolmström First one is undefined because there is no "ortnamn" with index 0 so the first one is empty – Alisso Aug 10 '12 at 16:47
    
@anders: because JS arrays start at 0, but sql ids generally start at 1. – Marc B Aug 10 '12 at 16:47
up vote 3 down vote accepted

The problem here is that json_encode($result); creates a JSON object, not an array. Hence jQuery.inArray isn't working.

Build a JavaScript array instead: [ "value", "value", "value", etc... ];

Or do something like this with the object:

var locations = {
    "Abårremöla" : 1,
    "Stuff" : 2,
    "Ludvika" : 1549
};

var query = "Ludvika";
if(query in locations) {
    var id = locations[query];
    alert(id); //displays 1549
}



For that to work you have to build up the object the other way around in PHP - so a PHP array with the location name as the indexer and the id as the value:

$query = mysql_query("SELECT ort_id, ortnamn FROM geo_orter");
while(($row = mysql_fetch_assoc($query))){
    $index = $row['ortnamn'];     //<- note!
    $result[$i] = $row['ort_id']; //<- note!
};
$allaOrterjson=json_encode($result);
share|improve this answer
    
But can I then control the index of each item? – Alisso Aug 10 '12 at 16:55
    
Each value has an id in the db so I'm trying to create a list with values and their ids as indexes so that I can look through it as the user chooses a value – Alisso Aug 10 '12 at 16:55
    
Or is there perhaps a way to look through the object instead? :) – Alisso Aug 10 '12 at 16:56
    
I understand the problem. What do you actually want to do with the object? Check if it has a certain value? – Anders Holmström Aug 10 '12 at 16:56
    
Updated my answer with an object-based solution. – Anders Holmström Aug 10 '12 at 16:59
up vote 0 down vote

Try this

var my_cars= new Array()
my_cars[0]="Mustang";

my_cars["family"]="Station Wagon";

my_cars["big"]="SUV";

for( var i in my_cars ){
    console.log( my_cars[i] );
}

var n = {
    "0": "Mustang",
    "family": " Station Wagon",
    "big": "SUV"
};
for( var i in n ){
    console.log( n[i] );
}
share|improve this answer

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