Question

我一直在谷歌搜索几个小时，还没有找到我一直收到的错误的答案。我有以下代码：

$varFirstName = $_POST['FirstName'];
$varPast = $varFirstName.' Past';
$varPresent = $varFirstName.' Present';
$varPastandPresent = $varFirstName.' Past and Present';
$CharacterID = "SELECT CharacterID FROM CHARINFO WHERE FirstName='$varFirstName'";

$sql="INSERT INTO CHARACTER_RELATIONS (RelationshipWith, CharacterID)
VALUES('$varPast', '$CharacterID'),
('$varPresent', '$CharacterID'),
('$varPastandPresent', '$CharacterID')";

mysqli_query($con,$sql)
or  die('Error: ' . mysql_error());

每次运行时我都会收到“错误：”。如果我使$ CharacterID = 49;然后它工作正常，所以我觉得“SELECT CharacterID FROM CHARINFO WHERE FirstName ='$ varFirstName'”有问题，但我无法弄清楚是什么。建议？

Answer 1

$ CharacterID =＆＃34; SELECT CHARID FROM CHARINFO WHERE firstName =＆＃39; $ varFirstName＆＃39;＆＃34 ;; //这会返回一个数组对象。所以你必须使用mysqli_fetch_array（）检索每一行的值，如下所示 // print_r（$ CharacterID）; exit（）; //检查数组值，同时你也执行了$ CharacterID。即mysqli_query（$ con，$ CharacterID）;

$varFirstName = $_POST['FirstName'];
$varPast = $varFirstName.' Past';
$varPresent = $varFirstName.' Present';
$varPastandPresent = $varFirstName.' Past and Present';
$Character= "SELECT CharacterID FROM CHARINFO WHERE FirstName='$varFirstName'";// this returns an array object
       $CharacterID= mysqli_query($con,$Character);
        while($row=mysqli_fetch_array($CharacterID)){
        $id=$row['CharacterID '];
        $sql="INSERT INTO CHARACTER_RELATIONS (RelationshipWith, CharacterID)
        VALUES('$varPast', '$id')";      
        mysqli_query($con,$sql)
        or  die('Error: ' . mysqli_error($con));  
}

变量中的变量中的SQL代码

1 个答案: