Using array in where clause in php

Question

I have an array $order_mealsInfo_ids which has the values like this

[{"meal_id":"33"},{"meal_id":"34"}]

Both the meal_id "33" and "34" has the same values(store_name,franchise_id,order_datetime) so I only want to get those values once that's why I am using distinct keyword.I want to use $order_mealsInfo_ids in my query,My query is :

    $get_franchise_info="Select distinct store_name,franchise_id,order_datetime from order_main where id 
IN (".implode(',',$order_mealsInfo_ids['meal_id']).")";

But its giving this error:

Notice: Undefined index: meal_id in C:\xampp\htdocs\learning\service\get_pending_orders_news.php on line 94

Warning: implode(): Invalid arguments passed in C:\xampp\htdocs\learning\service\get_pending_orders_news.php on line 94

mario.van.zadel · Accepted Answer · 2015-10-29 06:30:23Z

2

In your case $order_mealsInfo_ids is not an PHP array. It's a string containing a JSON array.

Please try this:

$order_mealsInfoJSON = '[{"meal_id":"33"},{"meal_id":"34"}]';
$order_mealsInfoArr = json_decode($order_mealsInfoJSON); // Convert JSON string to PHP array containing objects.

$order_mealsInfoIds = array();
foreach($order_mealsInfoArr as $order_mealsInfo) {
    $order_mealsInfoIds[] = $order_mealsInfo->meal_id;
}

$get_franchise_info="Select distinct store_name,franchise_id,order_datetime from order_main where id IN (".implode(',', $order_mealsInfoIds).")";

answered Oct 29, 2015 at 6:30

mario.van.zadel

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Parth Chavda · Accepted Answer · 2015-10-29 06:33:02Z

1

[{"meal_id":"33"},{"meal_id":"34"}] this is not Array this is json

you need to convert json to array

`$array = json_decode($json);
$mapedarray = array_map('current',$array);
$data = implode(',',$mapedarray);
echo $data;
`

so you get 33,34

`
 $get_franchise_info="Select distinct store_name,franchise_id,order_datetime from order_main where id 
IN (".implode(',',$mapedarray).")";
`

or

 `
     $get_franchise_info="Select distinct store_name,franchise_id,order_datetime from order_main where id 
    IN (".$data.")";
    `

answered Oct 29, 2015 at 6:33

Parth Chavda

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Comments

Ninju · Accepted Answer · 2015-10-29 06:33:53Z

0

Use this

    $data_jsn = '[{"meal_id":"33"},{"meal_id":"34"}]';
    $data = json_decode($data_jsn,true);
    $data_val = array_map(function($element) {
      return $element['meal_id'];
    }, $data);

    $get_franchise_info="Select distinct store_name,franchise_id,order_datetime from order_main where id 
    IN (".implode(',',$data_val).")";

answered Oct 29, 2015 at 6:33

Ninju

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