Question

我在MySQL中有一个UPDATE查询

UPDATE users 
SET fname='$fname',lname='$lname',username='$user',password='$pass',role='$role',status=$status 
where id=$id

当我在Phpmyadmin中通过手动赋值来运行它时，它运行得很好但是当我在PHP页面中使用它时它会给我一个错误消息。

Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

生成错误是因为我正在做echo mysql_error()我正在从这样的FORM中获取值;

$id=$_POST['id'];
$fname=$_POST['txt_fname'];
$lname=$_POST['txt_lname'];
$user=$_POST['txt_user'];
$pass=$_POST['txt_pass'];
$role=$_POST['txt_role'];
$status=$_POST['status'];

坦率地说，我听起来像一个愚蠢的问这么简单的问题，但我被困住了，需要一些帮助。感谢

表格HTML

<div id="cuntable" style=" width:400px; margin:0 auto;">
    <div class="CSSTableGenerator">
    <form name="userform" action="updateuser.php" method="post">
    <table >
    <tr>    
        <td colspan="2"> Edit / Modify User Details </td>
    </tr>
    <tr>
        <td>First Name </td>
        <td><input type="text" name="txt_fname" placeholder="<?php echo $row[1];?>" /></td>
    </tr>
    <tr>
        <td>Last Name </td>
        <td><input type="text" name="txt_lname" placeholder="<?php echo $row[2];?>" /></td>
    </tr>
    <tr>
        <td>Username</td>
        <td><input type="text" name="txt_user"  placeholder="<?php echo $row[3];?>"/></td>
    </tr>
    <tr>
        <td>Password</td>
        <td><input type="password" name="txt_pass"  placeholder="<?php echo $row[4];?>" /></td>
    </tr>
    <tr>
        <td>Username</td>
        <td>
        <select name="txt_role">
            <option value=""> Select User Role </option>
            <option value="user"> User </option>
            <option value="admin">Administrator</option>
            <option value="operator">Operator</option>
            <option value="accountant">Accountant </option>
        </select>
        </td>
    </tr>
    <tr>
        <td> Status </td>
        <td>
            <select name="status">
                <option value="1"> Active </option>
                <option value="0"> Suspended </option>
            </select>
        </td>
    </tr>
    </table>
    </div>
<br />
<input type="submit" value=" ADD USER " />
<input type="hidden" name="txt_id" value="<?php echo $row[0];?>"
</form>

MySQL：（PHP）UPDATE查询问题

0 个答案: