Question

我一直坚持这个问题几个小时，我研究的对我也没有帮助。我只是在输入与数据库中已有的名称匹配时才尝试更新我的数据库。如何修复此代码并感谢您的帮助。

         //Input 
      <form>
        <input  type="text" class="form-control" aria-describedby="emailHelp" name="name_input" placeholder="Enter Name">
              </div>

        <a href="" name="name_btn" class="btn btn-primary">Enter</a>
        </div> 
      </form>


    //The PHP code 

    $name = "";
    $name_input = "";
    $points = "";

    if(isset($_POST['name_btn'])){

        $name_input = strip_tags($_POST['name_input']);
        $name = $user['name'];
        $points = $user['points'];

    if($name_input == $name) {

        $points = points + 3;

    $upate_data = mysqli_query($con, "UPDATE users SET points='$points' WHERE id='4'");

    } 
    }

Answer 1

您可以使用select查询并获取返回行的计数来测试结果是否存在。例如;

$query = "SELECT * FROM $table WHERE username = '$name_input'";
$result = mysqli_query($conn, $query);
$count = mysqli_num_rows($result);
if($count > 0){
    //the result exists
    $points = points + 3;
    $upate_data = mysqli_query($con, "UPDATE users SET points='$points' WHERE id='4'");
}else{
    //result does not exist
}

此外，我看到您在此处获得name变量的$name_input = strip_tags($_POST['name']);变量name_input的数据，但输入字段的名称属性为<input type="text" class="form-control" aria-describedby="emailHelp" name="name_input" placeholder="Enter Name">的格式$('#members-search .typeahead').on("input", function(){ $('.tt-suggestion').on("click", changeFunction); });您应该检查出来。

Answer 2

你应该在你的表单中添加一个方法，同时href不适用于post，应该是这样的：

//Input 
<form method="post">
    <input  type="text" class="form-control" aria-describedby="emailHelp" name="name" placeholder="Enter Name">
    </div>

    <!-- <a href="" name="name_btn" class="btn btn-primary">Enter</a> -->

    <input type="submit" value="Submit" name="name_btn">
    </div> 
</form>


//The PHP code 

<?php

$databaseHost = 'localhost';
$databaseName = 'testing';
$databaseUsername = 'root';
$databasePassword = '';

$con = mysqli_connect($databaseHost, $databaseUsername, $databasePassword, $databaseName);

$name = "";
$name_input = "";
$points = "";

if(isset($_POST['name_btn'])){

    echo "aaa";

    $name_input = strip_tags($_POST['name']);
    $name = $user['name'];
    $points = $user['points'];


echo "&nbsp&nbsp Data updated successfully!";

    if($name_input == $name) {

    $points = points + 3;

    $upate_data = mysqli_query($con, "UPDATE users SET points='$points' WHERE id='4'");

    }
}

?>

如何在If语句中使用php正确更新mysql数据？

2 个答案: