升级后，php脚本不再工作 [英] php scripts no longer working after upgrade

问题描述

我最近从fedora core 5升级到cent os 4.4 with php 4.3

从mysql调用特定条目不再工作


如果我有一个链接display_story.php？id = 334它会知道显示

条目匹配id 334，我显示条目的代码看起来像

这个


$ Host =" localhost" ;;

$ User =" user";

$ Password =" password";

$ DBName =" database" ;;


$ Link = mysql_connect（$ Host，$ User，$ Password）;

$ Query =" SELECT * from table WHERE sid = $ id;

$ Result = mysql_db_query（$ DBName，$ Query，$ Link）;

while（$ Row = mysql_fetch_array（$ Result））{

print（" table width = \" 500 \"> \ n"）;

print（"< tr> \ n"）;

print（"< td valign = \&quo t; bottom \"> \ n"）;

print（"< p class = \" page_title \"> $ Row [title] \ n"）;

print（"< p> \ n"）;

print（"< i> $ Row [hometext]< / i> \ n"）;

print（"< / td> \ n"）;

print（"< td> \ n" ;）;

$ Topimage = $ Row [主题];

包含''top_image.php'';

print（"< ; / td> \ n"）;

print（"< / tr> \ n"）;

print（"< / table> \ n"）;

print（"< p> \ n"）;

print（" $ Row [bodytext] \ n"） ;

print（"< div align = right>< a href = \" javascript ：history.go（-1）\"> ;返回< /

a>< / div>< / td> \ n"）;

}

mysql_close（$链接）;

？>


我的日志中的错误消息告诉我$ Qu ery =" SELECT * from table

WHERE sid = $ id;是问题


[client 204.50.205.242] PHP注意：未定义变量：id in

**********

[client 204.50.205.242] PHP警告：mysql_fetch_array（）：提供

参数不是有效的mysql结果资源

****** *********************第10行


i试图启用register_globals但问题仍然存在


关于我如何在php.ini中解决这个问题的任何想法，而不必去和/或
写下我所有的脚本（今天我很懒）


谢谢

i recently upgraded from fedora core 5 to cent os 4.4 with php 4.3
mysql 4.1 and apache 2.0.52, and all of the php scripts i had which
called specific entries from mysql are no longer working

if i had a link display_story.php?id=334 it would know do display the
entry which matched id 334, my code to display the entry looks like
this

<?php
$Host="localhost";
$User="user";
$Password="password";
$DBName="database";

$Link=mysql_connect($Host, $User, $Password);
$Query="SELECT * from table WHERE sid=$id;
$Result=mysql_db_query($DBName,$Query, $Link);
while($Row=mysql_fetch_array($Result)){
print("<table width=\"500\">\n");
print("<tr>\n");
print("<td valign=\"bottom\">\n");
print("<p class=\"page_title\">$Row[title]\n");
print("<p>\n");
print("<i>$Row[hometext]</i>\n");
print("</td>\n");
print("<td>\n");
$Topimage=$Row[topic];
include ''top_image.php'';
print("</td>\n");
print("</tr>\n");
print("</table>\n");
print("<p>\n");
print("$Row[bodytext]\n");
print("<div align=right><a href=\"javascript: history.go(-1)\">Back</
a></div></td>\n");
}
mysql_close($Link);
?>

the error message in my logs tell me that $Query="SELECT * from table
WHERE sid=$id; is the problem

[client 204.50.205.242] PHP Notice: Undefined variable: id in
**********
[client 204.50.205.242] PHP Warning: mysql_fetch_array(): supplied
argument is not a valid mysql result resource in
*************************** on line 10

i have tried turning on register_globals but the problem persists

any ideas on how i can fix this in php.ini without having to go and re-
write all of my scripts (i''m kind of lazy today)

thanks

推荐答案

Host =" localhost";

Host="localhost";

User =" user";

User="user";

Password =" password";

Password="password";

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