Question

我正在尝试从外部json url将数据插入数据库。一些json在php中有空变量。如果变量中没有任何内容，我试图让它放入NULL或N / A.

这是我从阅读中想到的，但目前无法正常工作。

mysqli_query（）：第33行的update.php中的空查询

include('db.php');
 // create curl resource
    $ch = curl_init();

    // set url
    curl_setopt($ch, CURLOPT_URL, "bot.notenoughmods.com/1.8.9.json");

    //return the transfer as a string
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt($ch,CURLOPT_USERAGENT,'Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1.13) Gecko/20080311 Firefox/2.0.0.13');

    // $output contains the output string
  $json = json_decode(curl_exec($ch), true);
function isitempty($val){
    if (trim($val) === ''){$val = "NULL";}
    return $val;
}
foreach($json as $item) {

        $name = isitempty($item['name']);
        $version = isitempty($item['version']);
        $author = isitempty($item['author']);
        $link = isitempty($item['longurl']);
        $comment = isitempty($item['comment']);
        $repo = isitempty($item['repo']);
        $license = isitempty($item['license']);
        $time = isitempty($item['lastupdated']);
        foreach($item['dependencies'] as $dep) {
            $dep2 = isitempty($dep);
$query = "INSERT INTO mods (name,dependancies,version,author,link,repo,license,update_time) VALUES ('". $name."','". $dep2 ."','". $version."','". $author."','". $link."','". $repo ."','". $license ."','". $update_time."')";

$q = mysqli_query($con, $query1) or die (mysqli_error($con));
}
}

    // close curl resource to free up system resources
    curl_close($ch);

Answer 1

你有一个错字

$q = mysqli_query($con, $query) or die (mysqli_error($con));

不是$query1而是$query

当变量为空时，Mysql插入null

1 个答案: