将解码数据存储到不同的ID

Question

my code is :

//This is the data I am getting [{"x":1,"y":0,"width":2,"height":10},{"x":6,"y":0,"width":2,"height":9}] 

<?php
$position = json_decode($_POST['positionData'], true);
$select_id = "SELECT id FROM homegrid";
$select_id_exec = mysql_query($select_id);
//print_r($position[1]);
//print_r($position[2]); 
//getting result seperately
foreach ($position as $entry) {
    $x = $entry['x'];
    $y = $entry['y'];
    $width = $entry['width'];
    $height = $entry['height'];
    $positionjson = json_encode($entry);
    //print_r($positionjson);
    while($idFromDB = mysql_fetch_assoc($select_id_exec)) {
        //print_r($idFromDB);
        //echo $update = "update homegrid set position = '$positionjson' WHERE id = '" . $idFromDB['id'] . "' ";
        //mysql_query($update);
    }
}
?>

The output of update query is

update homegrid set position = '{"x":6,"y":0,"width":2,"height":9}' WHERE id = '7' 
update homegrid set position = '{"x":6,"y":0,"width":2,"height":9}' WHERE id = '8'

that means result of last position of array is getting.

How can i get result like

update homegrid set position = '{"x":1,"y":0,"width":2,"height":10}' WHERE id = '7' 
update homegrid set position = '{"x":6,"y":0,"width":2,"height":9}' WHERE id = '8'

?

My table structure is given below

id position
7 {"x":1,"y":0,"width":2,"height":10}
8 {"x":6,"y":0,"width":2,"height":9}

Could you please help me to solve this issue?

Answer 1

Try this code you dont need two loops

<?php
$position = json_decode($_POST['positionData'], true);
$select_id = "SELECT id FROM homegrid";
$select_id_exec = mysql_query($select_id);
//print_r($position[1]);
//print_r($position[2]); 
//getting result seperately
$i = 0;
while($idFromDB = mysql_fetch_assoc($select_id_exec)) {
    $x = $position[$i]['x'];
    $y = $position[$i]['y'];
    $width = $position[$i]['width'];
    $height = $position[$i]['height'];
    $positionjson = json_encode($position[$i]);
    //print_r($idFromDB);
    $update = "update homegrid set position = '$positionjson' WHERE id = '" . $idFromDB['id'] . "' ";
    mysql_query($update);
    $i++;
}
?>