我在這裏有一個代碼,其中它通過數據庫記錄......從不同的表中填充下拉列表。Mysqld意外終止,因爲一段PHP代碼
我有幾個表,其中1臺= 1個下拉列表
代碼:(showhide_dropdown)
<?php
$hostname = "localhost"; // usually is localhost, but if not sure,
check with your hosting company,
if you are with webune leave as localhost
$db_user = "root"; // change to your database password
$db_password = ""; // change to your database password
$database = "minquep_test"; // provide your database name
$db_table1 = "roles"; // leave this as is
$db_table2 = "companies";
$db_table3 = "albury_branch";
$db_table4 = "minquep_branch";
$db_table5 = "countries";
$db = mysql_connect($hostname, $db_user, $db_password);
mysql_select_db($database,$db);
?>
<?php
$roles_sql="SELECT role_id, role_name FROM $db_table1";
$comp_sql= "SELECT company_name FROM $db_table2";
$albury_sql= "SELECT albury_id, albury_name FROM $db_table3";
$minq_sql= "SELECT minquep_id, minquep_name FROM $db_table4";
$count_sql= "SELECT country_id, country_name FROM $db_table5";
$roles_result=mysql_query($roles_sql);
$comp_result=mysql_query($comp_sql);
$al_result=mysql_query($albury_sql);
$minq_result=mysql_query($minq_sql);
$count_result=mysql_query($count_sql);
$roles_options="";
$comp_options="";
$al_options="";
$minq_options="";
$count_options="";
while ($roles_row=mysql_fetch_array($roles_result)) {
$roles_id=$roles_row["role_id"];
$role=$roles_row["role_name"];
$roles_options.="<OPTION VALUE=\"$role\">".$role;
}
while ($comp_row=mysql_fetch_array($comp_result)) {
$company=$comp_row["company_name"];
$comp_options.="<OPTION VALUE=\"$company\">".$company;
}
while ($al_row=mysql_fetch_array($al_result)) {
$albury=$al_row["albury_name"];
$al_options.="<OPTION VALUE=\"$albury\">".$albury;
}
while ($minq_row=mysql_fetch_array($minq_result)) {
$minquep=$minq_row["minquep_name"];
$minq_options.="<OPTION VALUE=\"$minquep\">".$minquep;
}
while ($count_row=mysql_fetch_array($count_result)) {
$country=$count_row["country_name"];
$count_options.="<OPTION VALUE=\"$country\">".$country;
}
?>
首先,它確實工作得很好,但是當我打開網頁,其中的代碼被張貼...一個mysql.exe錯誤信息會突然彈出(正常的微軟錯誤)說:意外的錯誤...類似的東西。
然後一個MySQL錯誤/警告將我的網頁上顯示..
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
resource in C:\xampp\htdocs\minquep-native\pages\showhide_dropdown.php
on line 60
我不明白這個問題。我是PHP新手請幫忙。謝謝。
謝謝。我現在有另一個問題。我的phpmyadmin仍然拒絕訪問。嘖嘖。再次感謝。可能它的國家表不存在。但是,我上次運行我的程序時看到了它......它真的很奇怪。 – xirukitepe 2012-08-05 14:57:22