Question

这里发生了一些奇怪的事情，通常我不会有问题，或者我可能只是累了。但MySQL不接受我的语法将JSON插入表中。

foreach ($newArray as $k => $v) {
    $query = "SELECT `id` FROM `achievements` WHERE `achieveID`=".$v['ID']." LIMIT 1";
    if ($result = mysqli_query($default, $query)) {
        $row_cnt = mysqli_num_rows($result);
        $row_cnt > 0 ? $parse = false : $parse = true;
        mysqli_free_result($result);
    }   
    $v['required'] == 'None' ? $req = $v['required'] : $req = json_encode($v['required'], JSON_FORCE_OBJECT);
    if($parse) {                        
        $result = mysqli_query($default, "INSERT INTO `achievements` VALUES (NULL, '".$v['cat']."', '".$v['ID']."', '".$v['title']."', '".$v['type']."', '".$v['factionD']."', '".$v['factionE']."', '".$v['text']."', '".$v['doneText']."', '".$v['points']."', '".$v['number']."', ".$req.", '".$v['rewards']."')") or die ('Unable to execute query. '. mysqli_error($default));
    }
}

我的代码插入其他所有内容并执行其他行但是当它到达需要json数据的行时（即当$ v ['required']不等于none时）它不会插入它并给我错误。

Unable to execute query. You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '"0":"Into the Undertow","1":"Darkwater Abyss","2":"Primal Powers and Triton's Dr' at line 1

JSON（来自json编码）

{"0":"Into the Undertow","1":"Darkwater Abyss","2":"Primal Powers and Triton's Dread","3":"Stirring Interstellar Waves"}

似乎双引号引起了问题？

修改

我重写了我的代码！感谢您的提示！

$insert = $default->prepare("INSERT INTO achievements (catID, achieveID, achieveName, achieveType, factionDominion, factionExiles, achieveText, achieveCText, achievePoints, achieveNum, achieveReq, achieveRew) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)");
$insert->bind_param("iissiissiiss", $cat, $id, $title, $type, $factionD, $factionE, $text, $doneText, $points, $number, $required, $rewards);

foreach ($newArray as $k => $v) {
    $query = "SELECT id FROM achievements WHERE achieveID=".$v['ID']." LIMIT 1";
    if ($result = $default->query($query)) {
        $row_cnt = $result->num_rows;
        $row_cnt > 0 ? $parse = false : $parse = true;
    }   
    $v['required'] == 'None' ? $req = $v['required'] : $req = json_encode($v['required'], JSON_FORCE_OBJECT);
    if($parse) {    
        $cat = $v['cat'];
        $id = $v['ID'];
        $title = $v['title'];
        $type = $v['type'];
        $factionD = $v['factionD'];
        $factionE = $v['factionE'];
        $text = $v['text'];
        $doneText = $v['doneText'];
        $points = $v['points'];
        $number = $v['number'];
        $required = $req;
        $rewards = $v['rewards'];   
        $insert->execute();
    }
}
$default->close();

它的工作非常好，它也非常快，它在几秒钟内就增加了超过2000行！我很确定有一种更清晰的方式来编写它，但第一次使用prepare！

Answer 1

您的问题在'".$v['number']."', ".$req.",

$req变量包含字符串，您需要将其包装在单引号中

'".$v['number']."', '".$req."',

最好的选择是使用prepare语句来避免这些错误和SQL注入。

Answer 2

您需要转义引号和特殊字符。查看mysqli_real_escape_string

MySQL不会插入JSON

2 个答案: