如何使用提交按钮更新mysql数据库

Question

How do you update mysql database when i click on the update function. The fields that i want to update are in the textfield that i have created.

<script>

     function yourfunction() 
     {
         document.getElementById("name").readOnly = true;
         document.getElementById("number").readOnly = true;
         document.getElementById("status").readOnly = true;
         document.getElementById("expertise").readOnly = true;
         document.getElementById("remark").readOnly = true;

     }

    window.onload = yourfunction;

    function myfunction()
    {
        document.getElementById("name").readOnly = false;
        document.getElementById("number").readOnly = false;
        document.getElementById("status").readOnly = false;
        document.getElementById("expertise").readOnly = false;
        document.getElementById("remark").readOnly = false;
        document.getElementById("edit").style.visibility='hidden';
    }

So at first i am able to click on edit button so that my textfields can be edited. once i edited them, i can click on the update button to update those fields in mysql database.

    <?php
    $name = $_GET['name'];

    $connect = mysqli_connect("localhost", "root", "","test1");
    $output='';
    $sql = "SELECT Name, Number, Expertise, Status, Remarks FROM particulars WHERE Name ='". $name."'";

    $result = mysqli_query($connect, $sql);

    while($row= mysqli_fetch_array($result))
{
    $name = $row['Name'];
    $number =$row['Number'];
    $Expertise =$row['Expertise'];
    $status =$row['Status'];
    $remarks =$row['Remarks'];
} 

    echo "<label>Name</label> <input id = 'name' type='text' value='" .$name. "'/> <br>";
    echo "<label>Number</label><input id = 'number' type='text' value='" .$number. "'/><br>";
    echo "<label>Expertise</label><input id = 'expertise' type='text' value='" .$Expertise. "'/><br>";
    echo "<label>Status</label><input id = 'status' type='text' value='" .$status. "'/><br>";
    echo "<label>Remarks</label><input id = 'remark' type='text' value='" .$remarks. "'/><br>";
    ?>

<br><br><br>
<input type="button" id="edit" value = "Edit" onclick="myfunction()">
<input type="submit" id="update" value = "Update">

I would not mind using either a submit button or using ajax to solve this problem. Thank you very much!:)

Answer 1

First a note on your while loop, that passes through all all rows. In your case it should be just one. This works since there is one row only but you should use an approach that shouldn't use a while loop.

Back to your quest. You need a form (simplest way)

echo "<form action='mypage.php' method='POST'>"; // Start Form (Forms use names not ids btw)
echo "<label>Name</label> <input name = 'name' type='text' value='" .$name. "'/> <br>";
echo "<label>Number</label><input name = 'number' type='text' value='" .$number. "'/><br>";
echo "<label>Expertise</label><input name = 'expertise' type='text' value='" .$Expertise. "'/><br>";
echo "<label>Status</label><input name = 'status' type='text' value='" .$status. "'/><br>";
echo "<label>Remarks</label><input name = 'remark' type='text' value='" .$remarks. "'/><br>";
echo "<input type='submit' id='update' value = 'Update'>"; // This will send the data
echo "</form>"; // End form

mypage.php

$sql = "Update particulars  set Number = " .$_POST['number'] etc....

It is extremely ineffecient to have Name as a primary key. Numbers are alot easier to process not to mention multiple people can have same names.

Have Id as primary key and set it to auto_increment.

you can add it to your form like this

<input name = 'id' type='hidden' value='" .$id. "'/>

Answer 2

尝试这个：

UPDATE particulars SET Name = $name, Number = $number, Expertise = $expertise, Status = $status, Remarks = $remarks WHERE Name = $name;