Php 层叠下拉页面刷新
我一直在尽我所能让这个脚本工作，我只是不知道我做错了什么。我对PHP和Javascript非常陌生，我正在尝试修改一个在互联网上找到的脚本，以便在我的网站上获得级联下拉列表。这是一个相当简单的脚本，不幸的是它没有使用AJAX，所以需要刷新页面，但是如果我在这方面遇到困难，我认为我还不应该转向AJAX。我遇到的问题是，当页面刷新时，脚本应该使我从菜单中选择的内容保持不变，但它总是重置为“选择一”，这是默认设置。它在原始脚本中工作，但我做了很多修改，所以我不明白为什么我的不粘。原文是：Php 层叠下拉页面刷新,php,mysql,apache,xampp,Php,Mysql,Apache,Xampp,我一直在尽我所能让这个脚本工作，我只是不知道我做错了什么。我对PHP和Javascript非常陌生，我正在尝试修改一个在互联网上找到的脚本，以便在我的网站上获得级联下拉列表。这是一个相当简单的脚本，不幸的是它没有使用AJAX，所以需要刷新页面，但是如果我在这方面遇到困难，我认为我还不应该转向AJAX。我遇到的问题是，当页面刷新时，脚本应该使我从菜单中选择的内容保持不变，但它总是重置为“选择一”，这是默认设置。它在原始脚本中工作，但我做了很多修改，所以我不明白为什么我的不粘。原文是： <?p
<?php
//***************************************
// This is downloaded from www.plus2net.com //
/// You can distribute this code with the link to www.plus2net.com ///
// Please don't remove the link to www.plus2net.com ///
// This is for your learning only not for commercial use. ///////
//The author is not responsible for any type of loss or problem or damage on using this script.//
/// You can use it at your own risk. /////
//*****************************************
$dbservertype='mysql';
$servername='localhost';
// username and password to log onto db server
$dbusername='';
$dbpassword='';
// name of database
$dbname='dd';
////////////////////////////////////////
////// DONOT EDIT BELOW /////////
///////////////////////////////////////
connecttodb($servername,$dbname,$dbusername,$dbpassword);
function connecttodb($servername,$dbname,$dbuser,$dbpassword)
{
global $link;
$link=mysql_connect ("$servername","$dbuser","$dbpassword");
if(!$link){die("Could not connect to MySQL");}
mysql_select_db("$dbname",$link) or die ("could not open db".mysql_error());
}
//////// End of connecting to database ////////
?>
<!doctype html public "-//w3c//dtd html 3.2//en">
<html>
<head>
<title>Multiple drop down list box from plus2net</title>
<SCRIPT language=JavaScript>
function reload(form)
{
var val=form.cat.options[form.cat.options.selectedIndex].value;
self.location='dd.php?cat=' + val ;
}
</script>
</head>
<body>
<?
/*
If register_global is off in your server then after reloading of the page to get the value of cat from query string we have to take special care.
To read more on register_global visit.
http://www.plus2net.com/php_tutorial/register-globals.php
*/
@$cat=$_GET['cat']; // Use this line or below line if register_global is off
if(strlen($cat) > 0 and !is_numeric($cat)){ // to check if $cat is numeric data or not.
echo "Data Error";
exit;
}
//@$cat=$HTTP_GET_VARS['cat']; // Use this line or above line if register_global is off
///////// Getting the data from Mysql table for first list box//////////
$quer2=mysql_query("SELECT DISTINCT category,cat_id FROM category order by category");
///////////// End of query for first list box////////////
/////// for second drop down list we will check if category is selected else we will display all the subcategory/////
if(isset($cat) and strlen($cat) > 0){
$quer=mysql_query("SELECT DISTINCT subcategory FROM subcategory where cat_id=$cat order by subcategory");
}else{$quer=mysql_query("SELECT DISTINCT subcategory FROM subcategory order by subcategory"); }
////////// end of query for second subcategory drop down list box ///////////////////////////
echo "<form method=post name=f1 action='dd-check.php'>";
/// Add your form processing page address to action in above line. Example action=dd-check.php////
////////// Starting of first drop downlist /////////
echo "<select name='cat' onchange=\"reload(this.form)\"><option value=''>Select one</option>";
while($noticia2 = mysql_fetch_array($quer2)) {
if($noticia2['cat_id']==@$cat)
{echo "<option selected value='$noticia2[cat_id]'>$noticia2[category]</option>"."<BR>";}
else{echo "<option value='$noticia2[cat_id]'>$noticia2[category]</option>";}
}
echo "</select>";
////////////////// This will end the first drop down list ///////////
////////// Starting of second drop downlist /////////
echo "<select name='subcat'><option value=''>Select one</option>";
while($noticia = mysql_fetch_array($quer)) {
echo "<option value='$noticia[subcategory]'>$noticia[subcategory]</option>";
}
echo "</select>";
////////////////// This will end the second drop down list ///////////
//// Add your other form fields as needed here/////
echo "<input type=submit value=Submit>";
echo "</form>";
?>
<center><a href='http://www.plus2net.com'>PHP SQL HTML free tutorials and scripts</a></center>
</body>
</html>
<?php
$dbservertype='mysql';
$servername='localhost';
$dbusername='newuser';
$dbpassword='';
$dbname='supplydb';
connecttodb($servername,$dbname,$dbusername,$dbpassword);
function connecttodb($servername,$dbname,$dbuser,$dbpassword){
global $link;
$link=mysql_connect ("$servername","$dbuser","$dbpassword");
if(!$link){die("Could not connect to MySQL");}
mysql_select_db("$dbname",$link) or die ("could not open db".mysql_error());}
?>
<!doctype html public "-//w3c//dtd html 3.2//en">
<html>
<head>
<title>Multiple drop down list box from plus2net</title>
<SCRIPT language=JavaScript>
function reload(form){
var val=form.category.options[form.category.options.selectedIndex].text;
self.location='dd.php?cat=' + val ;
}
function reload2(form)
{
var val=form.cat.options[form.cat.options.selectedIndex].text;
var val2=form.subcat.options[form.subcat.options.selectedIndex].text;
self.location='dd.php?cat=' + val + '&cat2=' + val2 ;
console.log(val2);
}
</script>
</head>
<body>
<?
////// Grab Get after page refresh
$cat=$_GET['cat'];
$command="SELECT * FROM product_group WHERE type = 1";
$quer2 = mysql_query($command);
if(isset($cat) and strlen($cat) > 1){
$num = explode(' ',$cat);
$number = $num[0];
$strLen = strlen( $number );
for ( $i = 0; $i < $strLen; $i++ )
{$arr[] = $number{$i};}
$arr1 = $arr [0];
$arr2 = $arr [1];
$arr3 = $arr1.$arr2;
$command1="SELECT * FROM `product_group` WHERE `group` LIKE '".$arr3."%' AND type = '2'";
$quer = mysql_query($command1);
}
echo "<form method=post name=category action='dd-check.php'>";
///////////// Create First Drop Down Box
echo "<select name='category' onchange=\"reload(this.form)\"><option value=''>Select one</option>";
while($query = mysql_fetch_assoc($quer2)){
if($query['group']==$cat){
echo "<option selected value='$query[group]'>$query[group]</option>"."<BR>";}
else {echo "<option value='$query[group]'>$query[group]</option>";}
}
echo "</select>";
////////////// Create Second Drop Down Box
echo "<select name='subcat' onchange=\"reload2(this.form)\"><option value=''>Select one</option>";
while($query2 = mysql_fetch_array($quer)){
if($query2['group']==$cat2)
{echo "<option selected value='$query2[group]'>$query2[group]</option>"."<BR>";}
else {echo "<option value='$query2[group]'>$query2[group]</option>";}
}
echo "</select>";
echo "</form>";
print_r ($cat);
?>
</body>
</html>
找到了答案。很抱歉出现格式问题。从现在起，我将通过格式化程序运行所有代码。在原始代码中，他们调用的是一个我在数据库中甚至没有设置的ID字段。一旦我设置了字段并更正了代码，它就工作得很好。正确格式化代码将非常有帮助。这是您的第一个错误。抱歉，我将尝试立即修复此问题…使用空格而不是制表符，并确保每行前面至少有四个空格。顺便说一句，你不认为HTML3.2有点旧吗？请注意，在原始脚本中，select元素的名称是“cat”，其值是使用$_GET['cat']检索的，而在脚本中，你已经将select的名称更改为“category”，同时仍然尝试使用$_GET['cat']获取值。谢谢Yaniro。这是一个问题，另一个问题在我的答案中。在我的选择中，我正在查询组中的值和显示的内容。我需要一个id字段作为值。希望这对别人有帮助。