Question

我有这段代码。

$stmt4 = $conn->prepare("SELECT likedFour FROM UserData WHERE username = 'jim'");

现在，当username = jim时，这应该找到LikeFour行的值。

我有这个if语句。

if ($stmt4 == '') {
}

这不应该检查该值是否为空？它不起作用。

这是完整的代码。

$stmt = $conn->prepare("SELECT * FROM UserData WHERE username = ?");
$stmt->bind_param('s',$username);
//$username = $_POST["username"];
$username ="jim";
$stmt->execute();
$stmt->store_result();
$stmt1 = $conn->prepare("SELECT likedOne FROM UserData WHERE username = ?");
$stmt1->bind_param('s',$username);
//$username = $_POST["username"];
$username ="jim"; 


 echo "debug 2";
if ($stmt->num_rows == 0){ // username not taken
    echo "debug 2.5";
    die;

}else{
$result = mysqli_num_rows($stmt1);
echo "debug 2.7";
echo var_dump($stmt1);
    if ($stmt1 == 00000){
    echo "debug 3";
        $sql =  $conn->prepare("UPDATE UserData SET likedOne=? WHERE username=?");
        $sql->bind_param('ss',$TUsername,$Username);
//      $TUsername = $_POST["TUsername"];
//      $Username = $_POST["username"];
        $TUsername = "test";
        $Username = "jim";

    }
}

Answer 1

我会这样做：

$stmt4 = $conn->prepare("SELECT * FROM UserData WHERE username = 'jim'");
//This should grab the entire row where username == jim

然后

if(!$stmt4[likedFour]){
echo "Nothing has been found";
}

如果其他一切都很好，那应该是完美的。

Answer 2

尝试放置两个等号而不是一个。第一个等号表示您正在设置条件，而第二个等于表示您正在检查是否有东西等于第一个变量。

在PHP中查找变量的值

2 个答案: