Question

用户登录系统后，我正在进行用户个人资料更新。一切都很好，用户登录后，用户可以更改用户名传递等。但是有一个问题，当用户更改其详细信息时，它将更改整个用户数据库详细信息。这是我的编码

这是登录编码

<form class="modal-content animate" action="qcon.php" method="POST">
<div class="imgcontainer">
  <span onclick="document.getElementById('id01').style.display='none'" 
class="close" title="Close Modal">&times;</span>
  <img src="download.png" alt="Avatar" class="avatar">
</div>

<div class="container">
  <label><b>Username</b></label>
  <input type="text" placeholder="Enter Username" name="uname" required>

  <label><b>Password</b></label>
  <input type="password" placeholder="Enter Password" name="psw" required>

  <button type="submit">Login</button>
  <input type="checkbox" checked="checked" name ="submit"> Remember me
  </div>

这是连接数据库编码

 <?php


        $connect = new mysqli("localhost","root","","tenantsdb");

    $submit = isset($_POST['submit']);

    if($submit)
    {
        $uname = $_POST['uname'];
        $psw   = $_POST['psw'];
    }
 $get = mysqli_query($connect, "SELECT * FROM tenantsignup WHERE uname 
 ='$uname' AND psw ='$psw'");
 $get2 = mysqli_fetch_assoc($get);
 $id = $get2['id'];
 $num = mysqli_num_rows($get);

 if($num==1)
 {
    header("location:q1.php?id=$id");
 }

 else
 {
    echo"this username doesnt exsts";
 }
 ?>

以下是用户登录后的页面

<div class = "col-md-6">
<div class = "panel panel-default">
    <div class = "panel-body">

    <?php
    global $conn;
    $servername = "localhost";  //host name
    $username = "root"; //username
    $password = ""; //password
    $mysqli_database = "tenantsdb"; //database name

    //mysqli prepared statement 

    $conn = mysqli_connect($servername, $username, $password, 
    $mysqli_database) or die("Connection failed: " . 
    mysqli_connect_error());

    mysqli_select_db($conn,$mysqli_database) or die("Opps some thing went 
    wrong");
    $id = $_REQUEST['id'];

    $sql  = mysqli_query($conn,"SELECT * FROM tenantsignup WHERE id 
    ='$id'");
    $get3 = mysqli_fetch_assoc($sql);

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

        $uname = $_POST['uname'];
        $psw = $_POST['psw'];
        $name = $_POST['name'];
        $email = $_POST['email'];
        $contact_no = $_POST['contact_no'];
        $area = $_POST['area'];
        $gender = $_POST['gender'];
        $age = $_POST['age'];
        $max_budget = $_POST['max_budget'];
        $staying_with = $_POST['staying_with'];
        $race = $_POST['race'];

        $result = mysqli_query($conn,"UPDATE tenantsignup set 
        uname='$uname',psw='$psw',name='$name', 
        email='$email',contact_no='$contact_no',area='$area',
        gender='$gender',age='$age',max_budget='$max_budget',
        staying_with='$staying_with',race='$race'");

        if($result){

        echo 'Success!';

        }else{

        echo 'failed';

        }
    }
    ?>

    <form action ="q1.php" method="POST">
    <div class = "form-group">
        <label for ="id">ID</label>
        <input type="text" class ="form-control" id="id" name="id" value="<?
        php echo $id;?>">
    </div>
    <div class = "form-group">
        <label for ="uname">Username</label>
        <input type="text" class ="form-control" id="uname" 
         name="uname"value="<?php echo $uname;?>" >
    </div>
    <div class = "form-group">
        <label for ="psw">Password</label>
        <input type="text" class ="form-control" id="psw" name="psw"value="
        <?php echo $psw;?>" >
    </div>
    <div class = "form-group">
        <label for ="name">Name</label>
        <input type="text" class ="form-control" id="name" 
         name="name"value="<?php echo $name;?>" >
    </div>
    <div class = "form-group">
        <label for ="email">Email</label>
        <input type="text" class ="form-control" id="email" 
        name="email"value="<?php echo $email;?>" >
    </div>
    <div class = "form-group">
        <label for ="contact_no">Contact No</label>
        <input type="text" class ="form-control" id="contact_no" 
        name="contact_no"value="<?php echo $contact_no;?>" >
    </div>
    <div class = "form-group">
        <label for ="area">Area</label>
        <input type="text" class ="form-control" id="area" 
        name="area"value="<?php echo $area;?>" >
    </div>
    <div class = "form-group">
        <label for ="gender">Gender</label>
        <input type="text" class ="form-control" id="gender" 
        name="gender"value="<?php echo $gender;?>" >
    </div>
    <div class = "form-group">
        <label for ="age">Age</label>
        <input type="text" class ="form-control" id="age" name="age"value="
        <?php echo $age;?>" >
    </div>
    <div class = "form-group">
        <label for ="max_budget">Max Budget</label>
        <input type="text" class ="form-control" id="max_budget" 
    name="max_budget"value="<?php echo $max_budget;?>" >
    </div>
    <div class = "form-group">
        <label for ="staying_with">Staying With</label>
        <input type="text" class ="form-control" id="staying_with" 
    name="staying_with" value="<?php echo $staying_with;?>" >
    </div>
    <div class = "form-group">
        <label for ="race">Race</label>
        <input type="text" class ="form-control" id="race" 
    name="race"value="<?php echo $race;?>" >
    </div>

    <button type="submit" name="update">Update</button>
    </form>

Answer 1

您需要在UPDATE查询中添加WHERE子句：

 $result = mysqli_query($conn,"UPDATE tenantsignup set 
        uname='$uname',psw='$psw',name='$name', 
        email='$email',contact_no='$contact_no',area='$area',
        gender='$gender',age='$age',max_budget='$max_budget',
        staying_with='$staying_with',race='$race' WHERE id='$id'");

Answer 2

您需要在查询中添加 WHERE 。

用

替换您的查询

$result = mysqli_query($conn,"UPDATE tenantsignup set 
        uname='$uname',psw='$psw',name='$name', 
        email='$email',contact_no='$contact_no',area='$area',
        gender='$gender',age='$age',max_budget='$max_budget',
        staying_with='$staying_with',race='$race' WHERE id=$id");

注意：它不是更新数据的安全方式。你应该使用pdo或预备声明。

用户个人资料更新错误

2 个答案: