php - 变量内变量中的 SQL 代码

Question

我已经在谷歌上搜索了几个小时，但尚未找到我不断收到的错误的答案。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());

每次运行此命令时，我都会收到“错误:”。如果我使 $CharacterID=49;然后它工作得很好，所以我觉得“SELECT CharacterID FROM CHARINFO WHERE FirstName='$varFirstName'”有问题，但我不知道是什么。有建议吗？

Answer 1

$CharacterID= "从 CHARINFO 中选择字符 ID，其中firstName='$varFirstName'";//这返回一个数组对象。因此，您必须使用 mysqli_fetch_array() 检索每行的值，如下所示
//print_r($CharacterID);exit();//检查数组值，并且执行 $CharacterID。即 mysqli_query($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));  
}