简单的mysql代码没有运行，为什么？ [英] Simple mysql code doesn't run, why?

问题描述

您好，我已经投影了一个页面用于投票我的网站，我创建了一个包含2个INT字段（总票数和选民数）和一个id（工作）的mysql选项卡。现在我在运行此代码时遇到问题，为什么？

[PHP]<？php

$ user =" root";

$ password ="" ;;

$ database =" mydb";

$ db = mysql_connect（''localhost''，$ user，$ password）;

mysql_select_db（$ database）或死（无法选择数据库）;

$ sql =''SELECT * FROM`invest1` LIMIT 0,30'';

$ a = mysql_query（$ sql）或die（e）;

print" 1";

while（$ row = mysql_fetch_array（$ a））{

print" 2";

$ val = $ row [''val''];

$ part = $ row [''part''];

print $ val。" - "。$ part;

}

print" 3"

if（isset（$ _ GET [''vote'']））{

print" 4";

$ newval = $ val + $ _ GET [''vote''];

$ part = $ part + 1;

$ q =" UPDATE vote1 SET val =''$ newval''" ;;

$ a = mysql _query（$ q，$ db）;

$ q =" UPDATE vote1 SET part =''$ part''" ;;

$ a = mysql_query（$ q，$ db）;

echo"< a href = \" vote.php \">结果< / a>";

} else {

echo"媒体从0（ - ）到4（+）是：。$ val / $ part;

}

？> [/ PHP]

Hello, i have projected a page for voting my website, i have created a mysql tab with 2 INT fields (total votes and number of voters) and an id (working). Now I have problems running this code, why?
[PHP]<?php
$user="root";
$password="";
$database="mydb";
$db=mysql_connect(''localhost'',$user,$password);
mysql_select_db($database) or die( "Unable to select database");
$sql = ''SELECT * FROM `vote1` LIMIT 0, 30 '';
$a = mysql_query($sql) or die("e");
print "1";
while ($row = mysql_fetch_array($a)){
print "2";
$val=$row[''val''];
$part=$row[''part''];
print $val."--".$part;
}
print "3";
if (isset($_GET[''vote''])){
print "4";
$newval=$val+$_GET[''vote''];
$part=$part+1;
$q="UPDATE vote1 SET val = ''$newval''";
$a = mysql_query($q, $db);
$q="UPDATE vote1 SET part = ''$part''";
$a = mysql_query($q, $db);
echo "<a href=\"vote.php\">Results</a>";
}else{
echo "Media from 0( - ) to 4 ( + ) is: ".$val/$part;
}
?>[/PHP]

推荐答案

user =" root";

user="root";

password ="" ;;

password="";

database =" mydb";

database="mydb";

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