为什么我的mysqli_fetch_assoc不工作？

Question

我想回到$contact_id，但我得到的是：

success. 
the user id is 

例如，在我的insert.htm我输入$ CheckContact：

应该返回：the user id is 11。这是我的user表：

如果我在浏览器地址栏再次点击checkcontact.php然后我得到：

the user id is 23 

它总是23，无论什么$CheckContact是。 你能告诉我什么是错的吗？这里是我的代码：

<?php 

require('dbConnect.php'); 

$CheckContact = $_POST['phonenumber']; 
$sql = "SELECT * FROM user WHERE username = '$CheckContact'"; 
$result = mysqli_query($con, $sql); 
$check = mysqli_fetch_array($result); 
//if $CheckContact is in the user table... 
if(isset($check)) { 

    echo 'success.' . "<br>"; 

    // get the associated rows of $CheckContact 
     $row = mysqli_fetch_assoc($result); 
    // get the associated user_id in that row 

     $contact_id = $row["user_id"]; 
     echo "the user id is ", $contact_id; 

} 
//if $CheckContact is NOT in the user table... 
else { 

    echo 'failure'; 
    } 

?> 

Answer 1

why dont you use this: 
<?php 
$sql="Select query"; 
       $result=mysqli_query($con,$sql); 
       while($rws=mysqli_fetch_assoc($result)){ 
       ?> 
to print all data 
<?php 
} 
?> 

尝试空的（）为$检查这将指示变量是否包含空值或不

Answer 2

当我我的代码更改为：

<?php 

require('dbConnect.php'); 

$CheckContact = $_POST['phonenumber']; 
$sql = "SELECT * FROM user WHERE username = '$CheckContact'"; 
$result = mysqli_query($con, $sql); 
$num_rows = mysqli_num_rows($result); 

if($num_rows >= 1) { 

etc.... 

它工作正常。