在PHP中查找变量的值
我有这个密码在PHP中查找变量的值,php,mysql,Php,Mysql,我有这个密码 $stmt4 = $conn->prepare("SELECT likedFour FROM UserData WHERE username = 'jim'"); 现在,当username=jim时,这应该会找到像four这样的行的值 我有这个if语句 if ($stmt4 == '') { } 是否应该检查该值是否为空? 它不起作用了 这是完整的代码 $stmt = $conn->prepare("SELECT * FROM UserData WHERE user
$stmt4 = $conn->prepare("SELECT likedFour FROM UserData WHERE username = 'jim'");
现在,当username=jim时,这应该会找到像four这样的行的值
我有这个if语句
if ($stmt4 == '') {
}
是否应该检查该值是否为空?它不起作用了 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";
}
}
我会这样做:
$stmt4 = $conn->prepare("SELECT * FROM UserData WHERE username = 'jim'");
//This should grab the entire row where username == jim
然后
如果其他一切都很好,那么它应该工作得很好。尝试放置两个等号,而不是一个等号。第一个等号表示您正在设置条件,第二个等号表示您正在检查某个变量是否等于第一个变量。我放弃并删除了我的答案。越来越无聊了。你放弃我了吗?你没有得到一些脱节。我想你应该去看看mysqli准备语句的一些例子,然后再试一次。你需要发布发送查询并获取结果的整个代码块。好的。我现在就可以
if(!$stmt4[likedFour]){
echo "Nothing has been found";
}