Question

我有输入只为数据库输入类名数据库tbl_class

db_id  db_class    db_username db_pass
1      test
2     second test

第二个表单从这个

中读取数据库中的类信息

<?php
    $sql=mysqli_query($conn,"select * from tbl_class")or die(mysqli_error($conn));
    echo"<table class='ol-md-12 table-bordered table-striped table-condensed cf table-bordered'>";
    echo"<thead class='cf'>";
    echo"<tr>";
    echo"<th style='background:#f7ac01;font-size:13px;vertical-align: middle;text-align:center'rowspan='2' >ID</th>
    <th style='background:#f7ac01;font-size:13px;vertical-align: middle;text-align:center'rowspan='2' >Class</th>
    <th style='background:#f7ac01;font-size:13px;text-align:center;vertical-align: middle;text-align:center'rowspan='2' >Username</th>
    <th style='background:#f7ac01;font-size:13px;text-align:center;vertical-align: middle;text-align:center'rowspan='2' >Password</th>
    <th style='background:#f7ac01;font-size:13px;text-align:center;vertical-align: middle;text-align:center'rowspan='2' >Save</th>
    <th style='background:#f7ac01;font-size:13px;text-align:center;vertical-align: middle;text-align:center'rowspan='2' >Delete</th>";
    echo"</tr></thead><tbody>";
    while($row=mysqli_fetch_array($sql)){
        $id=$row['db_id'];
        $class=$row['db_class'];
        $user=$row['db_username'];
        $pass=$row['db_pass'];
        echo"<tr>";
        echo"<form method='post' action='";?><?php $_PHP_SELF ?><?php echo "'>";
        echo"<td data-title='Id'>"; echo $id;
        echo "<input type='text' name='txt_id' value='$id'>";echo"</td>";
        echo"<td data-title='Class'>"; 
        echo "<input type='text' name='class' value='$class'>"; echo"</td>";
        echo"<td data-title='Username'>"; 
        echo "<input type='text' name='txt_username' value='$user'>"; echo"</td>";
        echo"<td data-title='Password'>"; 
        echo "<input type='text' name='txt_pass' value='$pass'>"; echo"</td>";
        echo"<td data-title='Save'>"; 
        echo "<input type='submit' name='saveclass' value='Save' >";echo"</td>";
        echo"<td data-title='Delete'>"; 
        echo "<a href='companies.php?d=$id'><img src='../img/delete.png'></a>"; 
        echo"</td></form>";              
    }
    echo"</tr>";
    echo"</tbody></table>"; 
?>

之后，管理员可以设置用户名和密码，并使用此代码将其保存到数据库

if(isset($_POST['saveclass'])){
        $id=$_POST['txt_id'];
        $classe=strip_tags(mysqli_real_escape_string($conn,$_POST['class']));
        $user=strip_tags(mysqli_real_escape_string($conn,$_POST['txt_username']));
        $pass=strip_tags(mysqli_real_escape_string($conn,$_POST['txt_pass']));
        $query=mysqli_query($conn,"select db_username from tbl_class where db_username='$user' AND db_username <>''")or die(mysqli_error($conn));
        $count=mysqli_num_rows($query);echo $count;echo $id;echo $classe;
        if($count==0){
            $sql=mysqli_query($conn,"update tbl_class set db_class='$classe',db_username='$user',db_pass='$pass' where db_id='$id'")or die(mysqli_error($conn));
            header("location:companies.php?msg=2");
        }else{echo"This username exist please choose another one.";}

    }

问题是他给我一条消息，信息是更新它但是当我查看数据库时看不到更新信息

Answer 1

这里没有检查空格。如果您将用户名字段留空并提交表单，则每次都会提交表单。所以你的情况将是 -

if(isset($_POST['saveclass'])){
    $id=$_POST['txt_id'];
    $classe=strip_tags(mysqli_real_escape_string($conn,$_POST['class']));
    $user=strip_tags(mysqli_real_escape_string($conn,$_POST['txt_username']));
    $pass=strip_tags(mysqli_real_escape_string($conn,$_POST['txt_pass']));
    $query=mysqli_query($conn,"select db_username from tbl_class where db_username='$user'")or die(mysqli_error($conn));
    $count=mysqli_num_rows($query);echo $count;echo $id;echo $classe;
    if($count==0){
        $sql=mysqli_query($conn,"update tbl_class set db_class='$classe',db_username='$user',db_pass='$pass' where db_id='$id'")or die(mysqli_error($conn));
        header("location:class.php?msg=2");
    }else{echo"This username exist please choose another one.";}

}

在某些情况下更新查询不起作用

1 个答案: