php json_decode

Question

Test in some json code, like [{"a":"1"},{"a":"2"},{"a":"3"},{"b":"2"}], I want use json decode, get the data and make a judge, if first a nod is == b nod or fisrt a nod is != b nod. some code here.

<?php
header("Content-type: text/html; charset=utf-8");
$json = json_decode('[{"a":"1"},{"a":"2"},{"a":"3"},{"b":"2"}]',true);
$number=1;
foreach($json as $num){
if($num['a']!=$num['b']){
 if($num['a']){
 echo 'a'.$number.''.$num['a'].'<br />'; 
 }
}else{
 if($num['a']){
 echo 'b'.$number.''.$num['a'].'<br />'; 
 }
}
$number++;
}
?>

Now my result is:

a11 
a22 
a33

And I need get the result:

a11
b22
a33

• netcoder: You want to do what exactly? If first a node is == b node or fisrt a node is != b node, then what?
• cj333: @netcoder, I want make a foreach all the $num['a'] and make out which $num['a'] is equal to $num['b'] such as a11,b22,a33
• Xeoncross: "Test in some json code", sweet man.
• cj333: hey :) , test for further study.

Answer 1

How can you compare non-existing array fields?

there is no $num['b'] field, there are these fields, its array[4] (index 0..3)

• a:1
• a:2
• a:3
• b:2

$num in first iteration will hold

array("a"=>1);

so if you wanna compare to "b":2 you have to use this pattern:

$cmpr = array_shift($json);
...
if($num != $cmpr)

Array_shift : http://php.net/manual/en/function.array-shift.php

• Korvin Szanto: array_pop($json) will output an array. It needs to be array_shift(array_pop($json)) to actually work. Also, this will only work if you have b set in the final value.
• Marek Sebera: yeah, you're right. I've repaired my answer. And yes, this code works only on given data structure

Answer 2

You're trying to refer to the b variable when b is not currently defined. you must first loop through and find the b variable. http://codepad.org/0SnrPp6N

<?php
$json = json_decode('[{"a":"1"},{"a":"2"},{"a":"3"},{"b":"2"}]',true);
foreach ($json as $var) {
 if (isset($var['b'])) { $b = $var['b']; break; }
}
$i=0;
foreach ($json as $var) {
 if (!isset($var['a']))
 continue;
 if ($var['a']!=$b) 
 echo "a".++$i."$var[a]\n";
 else
 echo "b".++$i."$var[a]\n";
}

• Marek Sebera: not, if the position of B is always known (last element of array)
• Korvin Szanto: @MarekSebera You're correct, but he hasn't specified if it is or not. Because of that, we have to assume it isn't.
• Korvin Szanto: @cj333 this isn't complex at all, either of our answers will work correctly. Your problem is as I described, you're referring to the value of key b when key b isn't set.
• cj333: ok, both of u are right, give the mark to the first anwser and give you two +1, thanks all.
• Korvin Szanto: @cj333 if the b key is not always set in the last array, array_pop will be no use. If you are not aware of b's position, use my code.