JSON array as SQL WHERE argument in PHP

Question

So i have this json I got from a previous query:

{"HUID":"1","UIDS":"1,2,3","name":"home","type":"page","cat_id":"home"}

I print it like:

$SJON_String = json_encode($returnArray[0]);
echo $SJON_String;

Works, nice formatting.

Now I extract the UIDS string:

$JSON_Decoded = json_decode($SJON_String);
$JSON_UIDS = $JSON_Decoded->{'UIDS'}; 

echo "1st JSON UIDS: ".$JSON_UIDS."<BR>";

It prints nice string:

1,2,3

Now I want to use this string to give to my next query:

foreach ($JSON_UIDS as $UID_get)
    {

     echo "Now: ".$UID_get. "<BR>";

     $query2 = "SELECT * 
                FROM items
                WHERE UID LIKE '%" . $UID_get. "%' 
                ORDER BY name";

     if($result2 = $server->query($query2))
     {
       while ($row2 = $result2->fetch_assoc())
       {
        array_push($returnArray2, $row2);
        }
     }

It will give me an error: Warning: Invalid argument supplied for foreach(), because it's not a valid string object.

If I replace $JSON_UIDS with JSON_Decoded it will work, but that's not what I want.

I tried formatting the string in a bunch of different ways but I can't get it to accept the extracted string. I tried declaring as string() before, after, formatting it as "1,2,3", with quotes, going a little insane..

There must be a good and efficient way...

kitti · Accepted Answer · 2012-02-24 16:34:16Z

3

You're trying to loop through a string. Try turning the string into an array:

$UIDS = explode( ',', $JSON_UIDS );

foreach( $UIDS as $UID_get ) {
    ...

answered Feb 24, 2012 at 16:34

kitti

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Trey Copeland · Accepted Answer · 2012-02-24 16:38:53Z

1

You're trying to loop through "1,2,3" - You need to create some type of list or an array. Possibly using explode()

$UIDS = explode(',', $JSON_UIDS);

You can access them now at $UIDS[0], $UIDS[1], $UIDS[2]

answered Feb 24, 2012 at 16:38

Trey Copeland

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