mysql_fetch_array()期望参数1是给定的资源布尔值 [英] mysql_fetch_array() expects parameter 1 to be resource boolean given
问题描述
运行代码时出现此错误
mysql_fetch_array()期望参数1为资源布尔值。
< pre lang =xml> <? php
$ con = mysql_connect( localhost,username,password);
if(!$ con)
{
die(connectivity error.mysql_error());
}
mysql_select_db(cool,$ con);
if(isset($ _ POST ['ok']))
{
$ s = 从*中选择* ogin,其中username ='$ _ POST [uname]'和password ='$ _ POST [pwd]' ;
$ name = mysql_query($ s) 或 die(mysql_error());
$ i = 0;
while($ row = mysql_fetch_array($ name))
< span class =code-summarycomment> {
$ i ++;
}
if($ i > 0)
{
回波( HI);
}
其他
{
echo< script language =' javascript' > ;
echoalert('抱歉,你不能来找我');;
echo< / script > 跨度>;
}
}
?>
< html >
< head > < / head >
< body >
< 表格 操作 = 方法 = 发布 >
< 输入 type = text name = uname / >
< 输入 类型 = text name = pwd / >
< 输入 type = submit 名称 = ok value = 登录 / >
< span class =code-keyword>< / form >
< / body >
< / html >
< / body >
请给出正确的编码
con = < span class =code-keyword> mysql_connect( localhost,username,password);
if(!
< blockquote> con)
{
die(connectivity error.mysql_error()); < span class =code-summarycomment>
}
mysql_select_db(cool,
con);
如果(isset(
I am getting this error while i running the code
mysql_fetch_array() expects parameter 1 to be resource boolean given.
<?php
$con=mysql_connect("localhost","username","password");
if(!$con)
{
die("connectivity error".mysql_error());
}
mysql_select_db("cool",$con);
if(isset($_POST['ok']))
{
$s="select * from login where username='$_POST[uname]' and password='$_POST[pwd]'";
$name=mysql_query($s) or die(mysql_error());
$i=0;
while($row=mysql_fetch_array($name))
{
$i++;
}
if($i>0)
{
echo("hi");
}
else
{
echo "<script language='javascript'>";
echo "alert('sorry you cant come to me');";
echo "</script>";
}
}
?>
<html>
<head></head>
<body>
<form action="" method="post">
<input type="text" name="uname"/>
<input type="text" name="pwd"/>
<input type="submit" name="ok" value="Login"/>
</form>
</body>
</html>
</body>
pls give the correct coding
con=mysql_connect("localhost","username","password"); if(!
con) { die("connectivity error".mysql_error()); } mysql_select_db("cool",
con); if(isset(
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