Php 可变变量
这是我闹鬼的密码Php 可变变量,php,mysql,Php,Mysql,这是我闹鬼的密码 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["usernam
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)
如您所见,变量Type1、Type2、Type3、Type4、Type5都等于“empty”。
由于某些原因,如您所见,if语句不认为它是“空的”,因为它没有响应“debug 3”。什么。。。。。(也没有错误。)如果此代码
echo 'result is ' . $Type1, $Type2,$Type3,$Type4,$Type5;
从字面上产生这个输出
结果是空的
如果每个“空”之间都有空格,那么数据库中的类型*
值显然都是空的或空的或其他带有前导/尾随空格的组合
您将要与进行比较
trim($Type1) == 'empty'
此外,如评论中所述,从不将选择*
与绑定结果
结合使用。对于所选列以及它们的显示顺序,您应该始终明确,例如
SELECT id, dbUser, dbPassword, Type1, Type2, Type3, Type4, Type5 FROM ...
您没有在任何地方声明$Type1等。$stmt->bind_result($id、$dbUser、$dbPassword、$Type1、$Type2、$Type3、$Type4、$Type5);不是吗?如果是这样,为什么我可以打印它的值?你必须给这些变量赋值$Type1=“某物”//当然,直到您添加或删除列,或者您的数据库决定更改顺序
SELECT id, dbUser, dbPassword, Type1, Type2, Type3, Type4, Type5 FROM ...