PHP JSON解析给出错误 [英] PHP JSON parsing giving an error

问题描述

我正在尝试使用PHP解析JSON字符串，使用jQuery $.ajax以这种格式[{"value":"59"},{"value":"7"},{"value":"46"}]将JSON发送到PHP文件，但出于某些奇怪的原因，我一直收到此错误"Invalid argument supplied for foreach()"，在这里是我的PHP和jQuery代码，

I am trying to parse a JSON string using PHP, the JSON gets sent to the PHP file using jQuery $.ajax in this format, [{"value":"59"},{"value":"7"},{"value":"46"}] , but for some odd reason I keep getting this error "Invalid argument supplied for foreach()", here is my PHP and jQuery code,

jQuery:

    $(".systems").each( function(i, system) {
        // for each location block
        system = $(system);
        var sys = {
            'value' : $("select[data-prod='products']", system).val()
        };
        allSystems.push( sys );
    });

        $.ajax({
                type: 'POST',
                url: 'systems.php',
                dataType: 'json',
                data: { systems: JSON.stringify(allSystems), uid: uid },
                success: function(data){
                    alert(data)
                }
        });

PHP:

require_once 'classes/Zend/Json.php';

$json = $_POST['systems'];
$uid = $_POST['uid'];
$array= Zend_Json::decode("$json"); 

mysql_connect('localhost','user','pass') or die(mysql_error());
mysql_select_db('products') or die(mysql_error());

//insert the suppliers products into the database
foreach($array as $key){
    $query = mysql_query("INSERT INTO suppliersProducts (product_id, supplier_id) VALUES('".$key['value']."', '".$uid."' ) ") or die(mysql_error());
}

print_r($json);

如您所见，我正在使用Zend框架的JSON解码函数来解码传递给PHP的JSON字符串，而且该格式对我来说似乎是正确的，因此我不知道导致此错误的其他原因，也许是编码?有什么想法吗?

As you can see I am using the Zend framework's JSON decode function to decode the JSON string that gets passed to the PHP, and the format seems correct to me so I have no idea what else it could be that is causing this error, maybe encoding? Any ideas?

先谢谢！

推荐答案

看起来json_decode()无法处理该post参数中的数据.在脚本中添加一些错误处理.

Looks like json_decode() couldn't handle the data in that post parameter. Add some error handling to your script.

$test = json_decode($json, true);
if ( is_null($test) ) {
  die('data is not encoded as valid json');
}
else if ( !is_array($test) ) {
  die('Unexpected data structure. Array expected.');
}

您可能还对 json_last_error()

更新:也许是编码问题.由于您无权访问json_last_error()，因此您可能需要手动"检查字符串和编码，例如

update: Maybe it's an encoding issue. Since you don't have access to json_last_error() you might want to check the string and the encoding "manually", e.g.

if ( is_null($test) ) {
  // <--- for debugging purposes only
  echo '<pre>Debug: $json=';
  var_dump($json);
  echo 'hex=';
  for($i=0; $i<strlen($json); $i++) {
    printf('%02X ', ord($json[$i]));
  }
  echo '</pre>';
  // for debugging purposes only -->
  die('data is not encoded as valid json');

---

在将值作为字符串文字混合到sql语句中时，还必须使用 mysql_real_escape_string() ，请参见 CWE-89:在SQL命令("SQL注入" ')

---

You also have to use mysql_real_escape_string() when mixing the values as string literals into the sql statement, see CWE-89: Improper Neutralization of Special Elements used in an SQL Command ('SQL Injection')

$mysql = mysql_connect('localhost','user','password') or die(mysql_error());
mysql_select_db('products', $mysql) or die(mysql_error($mysql));

...

$sql = "
  INSERT INTO
    suppliersProducts
    (product_id, supplier_id)
  VALUES
    (
      '" . mysql_real_escape_string($key['value'], $mysql) . "'
      , '" . mysql_real_escape_string($uid, $mysql) . "'
    )
";
$query = mysql_query($sql, $mysql) or die(mysql_error($mysql));

甚至更好:使用预先准备好的参数化查询，请参见例如 PDO-准备好的语句和存储过程

Or even better: use prepared, parametrized queries, see e.g. PDO - Prepared statements and stored procedures

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