SQL 语法；检查与您的 MySQL 服务器版本相对应的手册，以获取在 '= '{search_term}'' 附近使用的正确语法

Question

我不能做实时搜索表的事情。有人能帮助我吗？ 这是我的代码。我只想显示我搜索过的数据...................................... ..................................................... ..................................................... ..................................................... ..................................................... …………

<?php

//include the connection file
include "conn.php";
$sql = "SELECT * FROM tblreservation";
if (isset($_POST['search'])) {
    $search_term = mysql_real_escape_string($_POST['search_box']);
        $sql .= "WHERE Name = '{search_term}'";
}

$query = mysql_query($sql) or die(mysql_error());
 ?>
 <form name="search_form" method="POST" action="trys.php" align="center">
 Search: <input type="text" name="search_box" value="" />
 <input type="submit" name="search" value="Search the table...">
 </form>
 <table width="70%" cellpadding="5" cellspace="5">

 <tr>
<td>ID</td>
<td>Name</td>
<td>Email</td>
<td>Packages</td>
      <td><select name="Packages" class="fieldsize">
      <option value="">select package</option>
      <option value="budget" <?php if($valid_Packages=='budget') echo     "selected='selected'";?>>Budget</option>
      <option value="standard" <?php if($valid_Packages=='standard') echo "selected='selected'";?>>Standard</option>
      <option value="super" <?php if($valid_Packages=='super') echo "selected='selected'";?>>Super</option>
      <option value="mega" <?php if($valid_Packages=='mega') echo "selected='selected'";?>>Mega</option>
    </select>
    <span class="err"><?php echo $error["Packages"];?></span></td>
</tr>
<td>Contactno</td>
<td>Gender</td>
      <td><input type="radio" name="gender" value="male" <?php     if($valid_gender=='male') echo "checked='checked'";?> />
    Male
    <input type="radio" name="gender" value="female" <?php if($valid_gender=='female')    echo "checked='checked'";?>/>
    Female <span class="err"><?php echo $error["gender"];?></span></td>
<td>file</td>
      <td><input type="file" name="file" value="upload" />
    <span class="err"><?php echo $error["file"];?></span></td>
<td>Address</td>
</tr>
<?php while ($row = mysql_fetch_array($query)) { ?>
<td><?php echo $row['id']; ?> </td>
<td><?php echo $row['Name']; ?> </td>
<td><?php echo $row['Email']; ?> </td>
<td><?php echo $row['Packages']; ?> </td>
<td><?php echo $row['Contactno']; ?> </td>
<td><?php echo $row['Gender']; ?> </td>
<td><?php echo $row['file']; ?> </td>
<td><?php echo $row['Address']; ?> </td>
</tr>
<?php } ?>

</table>

Answer 1

您在这一行缺少$ 和一个空格：

$sql .= "WHERE Name = '{search_term}'";

正确的行应该如下：

$sql .= " WHERE Name = '{$search_term}' ";

您当前生成的 SQL 语句正是这样的：

SELECT * FROM tblreservationWHERE Name = '{search_term}'

另外，我建议在你的 if 语句中检查 $_POST['search_box'] 而不是 $_POST['search'] 的存在，并且在附加它之前它实际上有一个值，因为这是你真正想要在查询中使用的值：

if (isset($_POST['search_box']) && $_POST['search_box']) {
    $search_term = mysql_real_escape_string($_POST['search_box']);
    $sql .= " WHERE Name = '{$search_term}' ";
}