Question

简而言之，我试图找出我的foreach声明有什么问题。我一直在努力寻找错误超过一天，而且我已经没时间了。该程序应该解析一个json数组并将其发布到mysqli数据库。

<?php

error_reporting(E_ALL);
ini_set('display_errors', 1);
$a = print_r(var_dump($GLOBALS),1);
echo htmlspecialchars($a);

$servername = "#";
$username = "#";
$password = "#";
$dbname = "#";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
echo "Connection Successful : ";

// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

// Read JSON file
$jsondata = file_get_contents('scripts/AUDIT_DIR/report.json');
echo "JSON File Read : ";

// Convert and Loop
$item = json_decode($jsondata, true);
echo "JSON File Decoded : ";

foreach($item as $arr)
{
    $id = $arr["id"];
    $hostname = $arr["hostname"];
    $ip = $arr["ip"];
    $package = $arr["package"];
    $publisher = $arr["publisher"];
    $origin = $arr["origin"];
    $version = $arr["version"];
    $size = $arr["size"];

    $sql = "INSERT INTO testtable(id, hostname, ip, package, publisher, origin, version, size)
    VALUES ('10', '$hostname', '$ip', '$package', '$publisher', '$origin', '$version', '$size')";

    if (mysqli_query($conn, $sql))
    {
        echo "New record created successfully : ";
    }
    else
    {
        echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
}

＆GT;

Answer 1

您json_decode()的回复可能无效，您可以在var_dump($item);

之后使用json_decode()查看此回复

如果json无法解码或者编码数据比递归限制更深，那么在php json_decode()中将返回NULL。 http://php.net/manual/en/function.json-decode.php

您需要妥善保护$item === null这样的情况，并且不要假设您将始终获得foreach()参数的有效回报。

显示$item = null时发生错误的示例 https://3v4l.org/oNr8P

PHP＆gt;为foreach（）提供的参数无效

1 个答案: