dongsu3654 2014-01-06 23:42
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PHP解析JSON字符串

I'm having great trouble passing some information from my Angularjs code to my PHP code for processing.

JS

Controller

var dataString = '{"round": {"number": 1,"drops": [{"pos": "0","cust": 1025}]}}';
dropService.updateDrops(dataString)
    .success(function(data) {
        console.log(data);
})

Service

updateDrops : function(drops) {
        return $http({
            url: "/app/php/update_rounds.php",
            method: "POST",
            data: drops,
            headers: {'Content-Type': 'application/x-www-form-urlencoded'}
        });
    }

PHP

$dropString = $_POST;

$dropArray = json_decode($_POST);
$a = $dropArray->{"round"}->{"number"};

echo "Round = ".$a;

What I expect to see on the console is

Round = 1

but what I get is

Round =

Obviously something is going screwy. However if I replace

$dropString = $_POST;

with

$dropString = '{"round": {"number": 1,"drops": [{"pos": "0","cust": 1025}]}}';

everything works out just fine so I know that the PHP will work if I can just get the right data to it. Where am I going wrong?

  • 写回答

2条回答 默认 最新

  • dongtaoxue4674 2014-01-06 23:47
    关注

    $_POST is an array of key=>value pairs

    You're looking for the entire POST BODY as a string. Use this:

    $dropString = http_get_request_body();
    

    Edit: If that doesn't work, use

    $dropString = file_get_contents('php://input');
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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