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

Answer 1

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";
}

Answer 2

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