的search.php

Question

我正在创建一个搜索功能或进程，从数据库中选择并在列表框中显示数据，用于选择已存在于MySQL数据库中的国家但我收到错误，我不知道如何修复它可以任何人帮助我我是新来的PHP。错误是（警告：mysql_fetch_array（）期望参数1是资源，布尔值在第36行的C：\ wamp \ www \ Unnamed Site 2 \ resources \ searchForm.php中给出）

的search.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connfig文件

Answer 1

如果查询失败，

mysql_query会返回FALSE。您需要检查返回的值。如果是FALSE，则可以使用mysql_error从数据库中获取错误消息。

Answer 2

你也可以在这里使用mysql_num_rows，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
     }
 }

您没有选择数据库使用此代码来选择db

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

将mysql_fetch_array更改为mysql_fetch_assoc。

php和搜索功能错误

的search.php

connfor_lamelchameltest.php

2 个答案: