Question

我在使用json_decode解决问题时遇到了麻烦。我得到了一个有效的json，它是通过$ _POST发送的。

这是传递的JSON对象，库存变量：

[{"item_name":"Screw Driver","item_desc":"asdasd","item_type":"weapon"},
{"item_name":"Brown Shoes","item_desc":"asdasd","item_type":"footwear"}]

这是html javascript代码。

<form action = "<?php echo $_SERVER['PHP_SELF'];?>" method = "post" accept-charset="UTF-8">
    <input type = "hidden" name = "inv" id = "inv"/>
    <input type = "submit" id = "btn_save" onclick = "saveInventory()" value = "Save"/>
</form>

<script type = "text/javascript">
function saveInventory(){
    var inv = document.getElementById("inv");
    inv.value = inventory;
}
</script>

这是将从输入类型隐藏

获取json的php脚本

<?php
if(isset($_POST['inv'])){
    $inv = $_POST['inv'];
    var_dump(json_decode($inv,true)); // returns NULL
}
?>

我一直在网上阅读很多json_decode问题但是大多数问题都有关于json_decode返回null的不同问题。有人可能会发现这里有什么问题吗？感谢。

Answer 1

您的inventory变量似乎是JavaScript对象数组，而不是JSON字符串。因此，您需要将其转换为JSON以将其存储在HTML输入值中：

function saveInventory(){
    var inv = document.getElementById("inv");
    inv.value = JSON.stringify(inventory); // convert to JSON string
}

Answer 2

<?php
    if(isset($_POST['inv'])){
        $json=json_decode($_POST['inv']);
        foreach($json as $j){
            echo "NAME: ".$j->{"item_name"};
            echo "<hr/>";
        }
    }
?>
<form method="post" action="<?php echo $_SERVER['SCRIPT_NAME'];?>">
    <input type="text" name='inv' value='[{"item_name":"Screw Driver","item_desc":"asdasd","item_type":"weapon"},{"item_name":"Brown Shoes","item_desc":"asdasd","item_type":"footwear"}]'/>
    <input type="submit" value="SEND"/>
</form>

json_decode在$ _POST上返回NULL

2 个答案: