php - 变量中变量中的SQL代码

我已经搜寻了几个小时，但仍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，我都会不断收到“错误：”。如果我使$ CharacterID = 49;然后它就可以正常工作，所以我觉得“从CHARINFO WHERE FirstName ='$ varFirstName'中选择字符ID”出了点问题，但我不知道是什么。有什么建议吗？

最佳答案

$ CharacterID =“从CHARINFO WHERE firstName ='$ varFirstName'中选择CharacterID”; //这将返回一个数组对象。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，然后执行$ 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));
}