分页代码中的错误 [英] error in pagination code
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问题描述
你好朋友,
我在我的分页脚本中遇到以下错误请帮我解决这个问题。
警告:mysql_fetch_array()期望参数1为资源,第16行的D:\ xampp \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ br />
以下是我的剧本
db.php
<?php
mysql_connect( localhost, root, );
mysql_select_db( library);
?>
data.php >
<?php
include(' ../ db.php');
if($ _ GET [' page'])
{
$ page = $ _ GET [' page'跨度>];
}
$ per_page_num = 2;
$ offset =($ page-1)* $ per_page_num ;
$ getdata =( select * from作者限制$ offset,1);
$ checking = mysql_query($ getdata);
// if($ checking === FALSE){
// die(mysql_error()); // TODO:更好的错误处理
// }
while ($ rows = mysql_fetch_array($ checking))
{
?> ;
<? = $ rows [' < span class =code-string> name'] ?> < < span class =code-leadattribute> br / >
<? php
}
?>
delt.php
<?php
include( ../ db.php);
$ query = mysql_query( select * from author);
$ count = mysql_num_rows($ query);
$ per_page = 1;
$ total = ceil($ count / $ per_page);
?>
< ul id = paginate >
<? php
for ($ i = 1; $ i< = $ total; $ i ++)
{
< span class =code-pagedirective>?>
< li id = <?= $ i?> > <? = $ i ?> < /利跨度> <跨度class =code-keyword>>
<? php
}
?>
< / ul >
< div < span class =code-attribute> id = content > < / div >
< div id = load > < / div >
< script src = ../ jquery.js 类型 = text / javascript > < / script >
< script >
$(document).ready(function(){
function display(){
$( #load)。fadeIn( slow);
$( #load)。html(' < img src =https://themarketingoak.files.wordpress.com/2015/07/circle-loading-animation.gif>' 跨度>);
}
function hide(){
$( #load)。fadeOut( 3000 );
}
$( #patchate li:first)。css( {color: red});
display();
$( #content)。load(' data.php?page = 1',hide());
$( #pateate li)。click(function(){
display();
$( #pateate li)。css({color : blue});
$( this )。css({color: red});
var num = $( this )。id;
$( #content)。load(' data.php?page =' + num,hide());
});
});
< / script >
解决方案
_GET [' page'])
{
page =
_GET [' page'];
}
hello friends,
I am facing following error in my pagination script please help me to solve this.
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in D:\xampp\htdocs\lib\author\data.php on line 16
following is my script
db.php
<?php
mysql_connect("localhost","root","");
mysql_select_db("library");
?>
data.php
<?php
include('../db.php');
if($_GET['page'])
{
$page =$_GET['page'];
}
$per_page_num=2;
$offset = ($page-1)* $per_page_num;
$getdata = ("select * from author limit $offset,1");
$checking = mysql_query($getdata);
//if($checking=== FALSE) {
// die(mysql_error()); // TODO: better error handling
//}
while ($rows = mysql_fetch_array($checking))
{
?>
<?=$rows['name']?><br/>
<?php
}
?>
delt.php
<?php
include("../db.php");
$query = mysql_query("select * from author");
$count = mysql_num_rows($query);
$per_page=1;
$total = ceil($count/$per_page);
?>
<ul id="paginate">
<?php
for ($i=1;$i<=$total;$i++)
{
?>
<li id="<?=$i?>"><?=$i?></li>
<?php
}
?>
</ul>
<div id="content"></div>
<div id="load"></div>
<script src="../jquery.js" type="text/javascript"></script>
<script>
$(document).ready(function(){
function display(){
$("#load").fadeIn("slow");
$("#load").html('<img src="https://themarketingoak.files.wordpress.com/2015/07/circle-loading-animation.gif">');
}
function hide(){
$("#load").fadeOut(3000);
}
$("#paginate li:first").css({color:"red"});
display();
$("#content").load('data.php?page=1',hide());
$("#paginate li").click(function(){
display();
$("#paginate li").css({color:"blue"});
$(this).css({color:"red"});
var num=$(this).id;
$("#content").load('data.php?page='+ num,hide());
});
});
</script>
解决方案
_GET['page']) {
page =
_GET['page']; }
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