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我正在创建一个搜索功能或流程以从数据库中进行选择并在 listdropbox 中显示数据，以选择 MySQL 数据库中已经存在的国家，但我遇到了一个错误，我不知道如何修复它任何人都可以帮助我我是 php 新手。错误是（警告：mysql_fetch_array() 期望参数 1 是资源，布尔值在 C:\wamp\www\Unnamed Site 2\resources\searchForm.php 第 36 行给出）

搜索.php

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Search page</title>
<link href="style/stylesheet.css" rel="stylesheet" type="text/css" />
<style type="text/css">
.searchtitle {
    text-align: center;
}
</style>
</head>

<body>

<!--<?php require_once('header.php'); ?>-->
<table width="80%" height="432">
  <tr>
    <td width="14%" valign="top"><!--<?php require_once('leftsideBar.php'); ?>-->
</td>
    <td width="86%" valign="top"><h2 class="searchtitle">Search Types</h2>
      <table width="95%" height="94">
      <form action="searchProcess.php" method="post" id="searchForm">

        <tr>
          <td width="20%"><label for="searchByName2">By Name</label>
            <select name="searchByName" id="searchByName2">
            </select></td>
          <td width="20%"><label for="searchByCountry">By Country</label>
            <select name="searchByCountry" id="searchByCountry">
             <option id="0">--select your country--</option>
             <?php
               require_once('Connections/connfor_lamelchameltest.php'); 

               $getallCountries = mysql_query("SELECT * FROM country");
               while($viewallCountries = mysql_fetch_array($getallCountries)){

                ?>

               <option id="<?php echo $viewallCountries['country_id']; ?> "><?php echo $viewallCountries['country_name'] ?></option>
                 <?php } ?>

            </select></td>
          <td width="25%"><label for="searchByGovernorate">By Governorate</label>
            <select name="searchByGovernorate" id="searchByGovernorate">
            </select></td>
          <td width="15%"><label for="searchByCity">By City</label>
            <select name="searchByCity" id="searchByCity">
            </select></td>
          <td width="20%"><label for="searchBySpecialization">By Specialization</label>
            <select name="searchBySpecialization" id="searchBySpecialization">
            </select></td>
        </tr>
        </form>
      </table>

  </tr>
</table>


</body>
</html>

connfor_lamelchameltest.php

<?php
$hostname_conn = "localhost";
$database_conn = "lam_el_chamel_db";
$username_conn = "root";
$password_conn = ".....";
$conn = mysql_pconnect($hostname_conn, $username_conn, $password_conn) or trigger_error(mysql_error(),E_USER_ERROR); 
?>

我添加我的配置文件

4

2 回答 2

3

FALSE如果查询失败，mysql_query返回。您需要检查返回值。如果是FALSE，您可以使用mysql_error从数据库中获取错误消息。

于 2013-02-27T06:41:17.093 回答
0

你也可以mysql_num_rows在这里使用，它会检查查询是否返回任何行。

 $getallCountries = mysql_query("SELECT * FROM country");
 if(mysql_num_rows($getallCountries) > 0) {
     while($viewallCountries = mysql_fetch_array($getallCountries)){
     // insert inside option
     }
 }

您尚未选择数据库使用此代码选择数据库

$conn = mysql_pconnect($hostname_conn, $username_conn, $password_conn) or trigger_error(mysql_error(),E_USER_ERROR);
mysql_select_db($database_conn, $conn);

更改mysql_fetch_array为mysql_fetch_assoc.

于 2013-02-27T06:43:34.557 回答