Question

这是我的HAUNTED代码。

ini_set("display_errors", true);
ini_set("html_errors", false); 
require "conn.php";
echo "debug 1";
$stmt2 = $conn->prepare("SELECT * FROM UserData WHERE username = ?");
$stmt2->bind_param('s',$username);
//$username = $_POST["username"];
$username ="netsgets";
$stmt2->execute();
$stmt2->store_result(); 
if ($stmt2->num_rows == 0){ // username not taken
echo "debug 2.5";

die;
}else{
// prepare query
$stmt=$conn->prepare("SELECT * FROM UserData WHERE username = ?");
// You only need to call bind_param once
$stmt->bind_param('s',$username);
$username = "netsgets";
// execute query
$stmt->execute(); 
$stmt->store_result();  
// bind variables to result
$stmt->bind_result($id,$dbUser,$dbPassword,$Type1,$Type2,$Type3,$Type4,$Type5);
//fetch the first result row, this pumps the result values in the bound variables

if($stmt->fetch()){
    echo 'result is ' . $Type1, $Type2,$Type3,$Type4,$Type5;
}


//var_dump($query2);
echo "hi";

echo "debug 2";



echo "debug 2.7";

    if ($Type1 == "empty"){

        echo "debug 3";
        $sql11 = $conn->prepare("UPDATE UserData SET likedOne=? WHERE username=?");
        $sql11->bind_param('ss',$TUsername,$Username);
//      $TUsername = $_POST["TUsername"];
//      $Username = $_POST["username"];
        $TUsername = "test";
        $Username = "netsgets";
        $sql11->execute();



    } 
}

这是它返回的内容（回声）。

Connected successfullydebug 1result is empty empty empty empty empty hidebug 2debug 2.7

如您所见，变量Type1，Type2，Type3，Type4，Type5都等于＆＃34;空＆＃34;。出于某种原因，正如你所看到的，if语句并不认为它是空的＆＃34;因为它没有回应＆＃34;调试3＆＃34;。什么.....（也没有错误。）

Answer 1

如果这段代码......

echo 'result is ' . $Type1, $Type2,$Type3,$Type4,$Type5;

从字面上产生这个输出......

结果为空空空空

每个＆＃34;空＆＃34;之间有空格，显然数据库中的Type*值都是"empty "或" empty"或其他一些与前导/尾随空格的组合

您想要与

进行比较

trim($Type1) == 'empty'

此外，正如评论中所述，从不将SELECT *与bind_result结合使用。您应该始终明确选择的列以及它们出现的顺序，例如

SELECT id, dbUser, dbPassword, Type1, Type2, Type3, Type4, Type5 FROM ...

改变变量

1 个答案: