Question

我正在为学生创建一个页面，他们可以通过三个不同的选项搜索他们的结果。我有问题。请指导我在哪里做错了

<form action="do_search.php" method="post">
<table width="540" border="0" align="center" cellpadding="0" cellspacing="0">
<tr>
  <td height="30"><strong>Search Criteria</strong></td>
  <td></td>
  <td><label for="criteria"></label>
    <select name="criteria" id="criteria">
      <option>Seach by Name</option>
      <option>Search by Father Name</option>
      <option>Search by Roll No.</option>
    </select></td>
</tr>
<tr>
  <td height="30">&nbsp;</td>
  <td></td>
  <td>&nbsp;</td>
</tr>
<tr>
  <td width="150" height="30"><strong>Enter Value:</strong></td>
  <td width="20"></td>
  <td width="370"><input name="value" type="text" id="value" size="40" /></td>
</tr>
<tr>
<td height="10" colspan="3"></td>
</tr>
<tr>
<td height="30"></td>
<td></td>
<td>
  <input name="submit" type="submit" value="Submit" id="submit_btn"/>
  <input name="reset" type="reset" value="Reset" id="reset_btn" />  </td>
</tr>
<tr>
<td colspan="3" height="10"></td>
</tr>
</table>
</form>

do.search.php

<?php
include 'authentication.php';
include 'includes/dbConnect.php';   
?>

<?php
    $criteria = $_POST['criteria'];
    $value = $_POST['value'];
    if ($value == '')
    {
        $myURL = 'error.php?eType=no_value';
        header('Location: '.$myURL);
        exit;
    }
    if ($criteria == "search by name")
    {
    $result = mysql_query( "SELECT * FROM students WHERE Name = '$value'" ) or die("SELECT Error: ".mysql_error());
    $num_rows = mysql_num_rows($result);
    }

    elseif ($criteria == "search by father name")
    {
    $result = mysql_query( "SELECT * FROM students WHERE Father_Name = '$value'" ) or die("SELECT Error: ".mysql_error());
    $num_rows = mysql_num_rows($result);
    }

    elseif ($criteria == "search by roll no.")
    $result = mysql_query( "SELECT * FROM students WHERE Roll_No. = '$value'" ) or die("SELECT Error: ".mysql_error());
    $num_rows = mysql_num_rows($result);
?>

请指导我在哪里做错了。

Answer 1

您的下拉值永远不会匹配，因为<option>Seach by Name</option>永远不会与$criteria == "search by name"匹配。

在您的表单中，我建议您执行以下操作：

<select name="criteria" id="criteria">
  <option value="1">Seach by Name</option>
  <option value="2">Search by Father Name</option>
  <option value="3">Search by Roll No.</option>
</select>

然后在PHP中，通常比较它们是否阻塞或切换。

switch($criteria) {
    case '1':
        // do something
    break;
    case '2':
        // do something
    break;
    case '3':
        // do something
    break;
}

旁注：我强烈建议切换到改进的版本API，即mysqli或使用PDO，并使用预准备语句。

粗略的例子（未经测试）：

if(isset($_POST['criteria'], $_POST['value'])) {
    $columns = array(1 => 'Name', 2, 'Father_Name', 3, 'Rol_No.');
    $db = new mysqli('localhost', 'username', 'password', 'database');  
    $criteria = $_POST['criteria'];
    if(!isset($columns[$criteria])) {
        // column not found
        exit;
    }
    $column_selected = $columns[$criteria];
    $value = '%' . $_POST['value'] . '%';
    $select = $db->prepare("SELECT * FROM students WHERE $column_selected LIKE ? ");
    $select->bind_param('s', $value);
    $select->execute();
    if($select->num_rows > 0) {
        // found
    } else {
        // none found
    }
}

Answer 2

首先，兄弟$_POST['criteria']获取空值无论何时编写搜索查询，请始终使用LIKE子句。

这是解决方案Rizwan sahib：

<select name="criteria" id="criteria">
      <option name="name">Seach by Name</option>
      <option name="father">Search by Father Name</option>
      <option name"roll">Search by Roll No.</option>
</select>


<?php
    $result = '';
    if ($criteria == "name")
    {
    $result = mysql_query( "SELECT * FROM students WHERE Name LIKE '%$value'%" ) or die("SELECT Error: ".mysql_error());
    $num_rows = mysql_num_rows($result);
    }

    elseif ($criteria == "father")
    {
    $result = mysql_query( "SELECT * FROM students WHERE Father_Name LIKE '%$value%'" ) or die("SELECT Error: ".mysql_error());
    $num_rows = mysql_num_rows($result);
    }

    elseif ($criteria == "roll"){
    $result = mysql_query( "SELECT * FROM students WHERE Roll_No. LIKE '%$value%'" ) or die("SELECT Error: ".mysql_error());
    $num_rows = mysql_num_rows($result);
}
$row = mysql_fetch_array($result);
?>

“％”符号用于在模式之前和之后定义通配符（缺少字母）。

希望您找到解决方案

在搜索结果有困难的PHP

2 个答案: