PHP Switch案例问题 [英] PHP Switch Case Problem
本文介绍了PHP Switch案例问题的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我正在尝试使用不同的按钮填充列表框。第一个工作正常，当我点击它时，列表框填充。问题是其他四个选项。我需要更改MySQL查询以反映按钮，以便列表框正常填充。
I am trying to populate a listbox with different buttons. The first one works fine, when I click on it the listbox populates. The problem is the other four options. I need to change the MySQL query to reflect the button so that the listbox populates proerly.
<?php
$dbc = mysql_connect('*','*','*')
or die('Error connecting to MySQL server.');
mysql_select_db('*');
$result = mysql_query("select * from tblRestaurants order by RestName ASC");
while ($nt= mysql_fetch_assoc($result))
$arrData[] = $nt;
if(isset($_GET["ajax"]))
{
echo json_encode($arrData);
die();
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.0/jquery.min.js" type="text/javascript"></script>
<script type="text/javascript">
function displayPlace()
{
$.getJSON("index.php?ajax=true", function(data) {
$.each(data, function(index, objRecord) {
var option=document.createElement("option");
option.value=objRecord.RestID;
option.text=objRecord.RestName;
$("#Doggie").append('<option value="' + objRecord.RestID + '">' + objRecord.RestName + '</option>');
});
});
}
function displayCuisine()
{
$.getJSON("index.php?ajax=true", function(data) {
$.each(data, function(index, objRecord) {
var option=document.createElement("option");
option.value=objRecord.CuisineID;
option.text=objRecord.CuisineName;
$("#Doggie").append('<option value="' + objRecord.RestID + '">' + objRecord.RestName + '</option>');
});
});
}
</script>
<title>SEARCH</title>
</head>
<body>
<form>
<button type="button" onclick="javascript:displayPlace();">Place</button>
<button type="button" onclick="javascript:displayCuisine();">Cuisine</button>
<button type="button" onclick="javascript:displayCity();">City</button>
<button type="button" onclick="javascript:displayState();">State</button>
<button type="button" onclick="javascript:displayZipCode();">Area</button>
<br />
<select name="Doggie" id="Doggie"></select>
<br />
</form>
</body>
</html>
推荐答案
dbc = mysql_connect（' *'，' *'，' *' ）
或 die（' 连接MySQL服务器时出错。'）;
mysql_select_db（' *'）;
dbc = mysql_connect('*','*','*') or die('Error connecting to MySQL server.'); mysql_select_db('*');
result = mysql_query（ 从tblRestaurants中选择*由RestName ASC命令）;
while （
result = mysql_query("select * from tblRestaurants order by RestName ASC"); while (
nt = mysql_fetch_assoc（
nt= mysql_fetch_assoc(
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