PHP Javascript Mysql插入

Question

我正在尝试使用java-script提示符将值插入到MySQL中。我有一个html文件和一个php文件。

另存为Test.html的html文件

 <button onclick="myFunction()">Create Project</button>

<script>
  function myFunction() {
      var project = prompt("Please enter project name");
      if (project != null && project !="") {
          $.post("conn.php", { project : project });
      }
  }
  </script>

The conn.php

<?php
$servername = "localhost";
$username = "root";
$password = "Password";
$dbname = "db1";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$proj = $_POST['project'];
$sql = "INSERT INTO projects (project_name) VALUES ('$proj')";


if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
?> 

conn.php可以很好地工作，如果我删除post参数，它会插入值。

例如，而不是这样：

$proj = $_POST['project'];
$sql = "INSERT INTO projects (project_name) VALUES ('$proj')";

如果我使用：

$proj = "NewProject";
$sql = "INSERT INTO projects (project_name) VALUES ('$proj')";

NewProject被插入到数据库中。

我不确定我的index.html中是否遗漏了什么，它没有将提示符中的值发布到php脚本中。我尝试了echo $_POST['project'];而不是插入到MySQL中。回声丢失。

Answer 1

我已经运行了你给定的代码，它只运行我在脚本代码上面添加了jquery链接

请检查此更正，

<button onclick="myFunction()">Create Project</button>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script>
  function myFunction() {
      var project = prompt("Please enter project name");
      if (project != null && project !="") {
          $.post("conn.php", { project : project },function(response){
             console.log(response);
         });
      }
  }
  </script>

我还在conn.php文件中添加了带$_POST参数的isset()函数

<?php
$servername = "localhost";
$username = "root";
$password = "Password";
$dbname = "db1";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$proj = isset($_POST['project'])?$_POST['project']:'';
$sql = "INSERT INTO projects (project_name) VALUES ('$proj')";


if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
?>