Question

我有一个包含5列的表格：ID，姓名，电子邮件，评论＆amp;约会时间。我创建了一个搜索表单，让用户通过姓名或电子邮件搜索记录。

这是我的代码：

<?php

// database info
$host = "localhost";
$username = "root";
$password = "root";
$db_name = "test";
$tbl_name = "abctable";

// conntect to server and select database
mysql_connect("$host", "$username", "$password") or die ("cannot connect server");
mysql_select_db("$db_name") or die ("cannot select DB");

// get data from search form
$name=$_POST['name'];
$email=$_POST['email'];

// if name & email is not empty, do this
if ((!empty($name)) || (!empty($email))) {
    if(isset($_REQUEST['submit'])){
    $sql=" SELECT * FROM $tbl_name WHERE name like '%".$name."%' OR email ='".$email."' ";
    $q=mysql_query($sql);
    }
    else{
        $sql="SELECT * FROM $tbl_name";
        $q=mysql_query($sql);
    }
}
else{
    echo "Please type something in the search box below:<br><br>";
}
?>

<!-- Search Form -->
<form method="post" action="search.php">
<table width="200" border="1">
  <tr>
    <td>Name</td>
    <td><input type="text" name="name" /></td>
    <td>Email</td>
    <td><input type="text" name="email" /></td>
    <td><input type="submit" name="submit" value=" Find " /></td>
  </tr>
</table>
</form>

<?php 
// Show this Result Table when Search Form is not empty
if ((!empty($name)) || (!empty($email))) { ?>
    <!-- Result Table -->
    <table border="1">
    <tr>
        <td>ID</td>
        <td>Name</td>
        <td>Email</td>
        <td>Comment</td>
        <td>Date &amp; Time</td>
    </tr>
    <?php while($res=mysql_fetch_array($q)){ ?>
    <tr>
        <td><?php echo $res['id'];?></td>
        <td><?php echo $res['name'];?></td>
        <td><?php echo $res['email'];?></td>
        <td><?php echo $res['comment'];?></td>
        <td><?php echo $res['datetime'];?></td>
    </tr>
    <?php } ?>
    </table>
<?php } ?>

<?php mysql_close(); // close database ?>

我遇到了电子邮件搜索SQL的问题。该名称正在运行并成功返回正确的记录。如何解决这个问题？

Answer 1

可能会将查询更改为

$sql="SELECT * FROM $tbl_name 
WHERE (".$name." <> '' and name like '%".$name."%')
  OR (".$email." <> '' and email like '%".$email."%')."' ";

如果变量为空，这将排除该变量。

Answer 2

保持清洁这样做：

if(isset($_REQUEST['submit']))
{

$sql=" SELECT * FROM $tbl_name WHERE name like $name OR email like $email
$q=mysql_query($sql);

}else
{

echo "Please type something in the search box below:<br><br>";

}

我相信你做的事情会更好。我有一个登录脚本，如果以上操作不起作用，我会检查并发布。

SQL：OR语句不起作用

2 个答案: