php json decode result in another json decode - php

Question

I have 2 json data from nosql. first match if the search word match in $array1, get the item number then put item number into $array2, get the price of custom search require. But my code cause Invalid argument supplied for foreach() in
foreach($json2[$num] as $data2)
$str = 'paper';
$array1 = '[{"a":"1","b":"book"},{"a":"2","b":"paper"}]';
$array2 = '[{"1":["17.00","22.00"]},{"2",["4.50","6.00"]}]';
$json1 = json_decode($array1);
$json2 = json_decode($array2,true);
foreach($json1 as $data1){
if(preg_match('#'.$data1->b.'#',$str,$match)){
$num = $data1->a; // $num = 2
}
}
foreach($json2[$num] as $data2){
foreach($data2 as $newdata){
echo $newdata.'<br />'; // 4.50, 6.00
}
}

Answer 1

First off your JSON for $array2 is not valid. It should be:
[{"1":["17.00","22.00"]},{"2":["4.50","6.00"]}]
-----------------------------^
This should be a ":", not a ","
Your JSON is actually an array of objects (arrays). var_dump($json2); shows this.
array(2) {
[0]=>
array(1) {
[1]=>
array(2) {
[0]=>
string(5) "17.00"
[1]=>
string(5) "22.00"
}
}
[1]=>
array(1) {
[2]=>
array(2) {
[0]=>
string(4) "4.50"
[1]=>
string(4) "6.00"
}
}
}
You're gonna need to loop over it like this:
foreach($json2 as $data2){
if(array_key_exists($num, $data2)){
$data2 = $data2[$num];
foreach($data2 as $newdata){
echo $newdata.'<br />'; // 4.50, 6.00
}
}
}

Answer 2

If $num = 2 in your comments is correct, you'll be accessing the third element in $json2 but you can't since there only are two.
Update
Oops, how did I miss this? Your $array2 already has the indices right there, you're just not loading them properly. You can simply loop through $array2 and look for the key. However, a better solution would be to load the data properly, by filling $array2 as a dictionary rather than a list.