Question

我希望通过使用函数单击插入和更新按钮来插入和更新客户数据，但是当我单击插入或更新按钮时，没有插入或更新数据是什么问题我的代码？

<form action="Customer.php" method="post">

<div>

<form>

Phone No <input type="number"  placeholder="Search"  name="phoneno" />

First Name <input type="text" name="FirstName" />

Last Name<input type="text" name="LastName" />

Phone No<input type="number" name="CustomerPhoneNo" />

Address<input type="text" name="Address" />

Customer ID<input type="number" name="CustomerID" />

<input type="Submit"  value="Add Customer"  style="font-size:20px"  onClick="insert()"> 

<input type="Submit"  value="Update Customer"  style="font-size:20px" onClick="update()"> 

 </form>

 </div>


"Customer.php" 

<?php

$dbhost="127.0.0.1";
$dbname="root";
$dbuser="info";
$dbpsd="";

$link = mysqli_connect("$dbhost", "$dbuser", "$dbpsd", "$dbname");


if($link === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
}

$phoneno = mysqli_real_escape_string($link, $_POST['phoneno']);
$FirstName = mysqli_real_escape_string($link, $_POST['FirstName']);
$LastName = mysqli_real_escape_string($link, $_POST['LastName']);
 $CustomerPhoneNo = mysqli_real_escape_string($link, $_POST['CustomerPhoneNo']);
 $Address=mysqli_real_escape_string($link, $_POST['Address']);
 $CustomerID = mysqli_real_escape_string($link, $_POST['CustomerID']);


 function insert(){

     $sql = "INSERT INTO clientinfo(phoneno, FirstName, LastName,CustomerPhoneNo,Address,CustomerID) VALUES ('$phoneno', '$FirstName', '$LastName','$CustomerPhoneNo','$Address','$CustomerID')";
echo "<span>Data Inserted successfully...!!</span>";

  if(mysqli_query($link, $sql)){
    echo "Records added successfully.";
} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

 }

 function update(){


 $sql="UPDATE clientinfo SET FirstName='$FirstName', LastName='$LastName',Address='$Address',CustomerID='$CustomerID WHERE phoneno='$phoneno' ";



if (mysqli_query($link, $sql)) {
    echo "Record updated successfully";
} else {
    echo "Error updating record: " . mysqli_error($link);
}
 }
 close connection
mysqli_close($link);
?>

Answer 1

试试这个：

删除第二个标记并将第一个标记放在该位置。

其他方面，您需要查看代码。

查看本教程 PHP Tutorial

Answer 2

您不能使用onclick标签来确定所请求的页面PHP功能一个简单的方法。在customer.php文件中替换以下代码。

替换：

function insert()

with：

if($_POST['Submit'] == 'Add Customer')

并替换：

function update()

with：

if($_POST['Submit'] == 'Update Customer')

Answer 3

你正在调用像js函数这样的php函数。他们不这样做。将它们定义为 -

function insert($link, $post){

    $phoneno = mysqli_real_escape_string($link, $post['phoneno']);
    $FirstName = mysqli_real_escape_string($link, $post['FirstName']);
    $LastName = mysqli_real_escape_string($link, $post['LastName']);
    $CustomerPhoneNo = mysqli_real_escape_string($link, $post['CustomerPhoneNo']);
    $Address=mysqli_real_escape_string($link, $post['Address']);
    $CustomerID = mysqli_real_escape_string($link, $post['CustomerID']);

    $sql = "INSERT INTO clientinfo(phoneno, FirstName, LastName,CustomerPhoneNo,Address,CustomerID) VALUES ('$phoneno', '$FirstName', '$LastName','$CustomerPhoneNo','$Address','$CustomerID')";
echo "<span>Data Inserted successfully...!!</span>";

  if(mysqli_query($link, $sql)){
      echo "Records added successfully.";
  } else{
      echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
  }
}

并将它们称为 -

insert($link, $_POST); // if it is inserting

使用php mysql插入和更新数据库

3 个答案: