我有一个非常简单的问题,即使用php/sql发送多个查询并将每个结果添加到关联数组中。与PHP/SQL关联数组中的多个SQL查询
基本上,我已经在关联数组中分配了每个查询。
每个结果对每个查询进入我的多维数组$el['nameofthequery']['resultsofthequery']
//connection information
$host = "localhost";
$user = "root";
$password = "";
$database = "sghi";
$el = array();
$query = array();
//make connection
$server = mysql_connect($host, $user, $password) or die('Could not connect to mysql server.');
$connection = mysql_select_db($database, $server);
//query the database
$query['mylist1'] = mysql_query("SELECT CompanyName AS label,ContactName AS value FROM suppliers") or die('Could not select database.');
$query['mylist2'] = mysql_query("SELECT ContactTitle AS label,City AS value FROM suppliers") or die('Could not select database.');
//build array of results
// Check if there is any results
if(mysql_num_rows($query['mylist1']) == 0) {
echo "No results found";
exit; // so exit
}
else {
while($row = mysql_fetch_assoc($query['mylist1'])){
$el['mylist1'][] = $row;
}
}
// Check if there is any results
if(mysql_num_rows($query['mylist2']) == 0) {
echo "No results found";
exit; // so exit
}
else {
while($row = mysql_fetch_assoc($query['mylist2'])){
$el['mylist2'][] = $row;
}
}
//echo JSON to page
$response = json_encode($el);
echo $response;
mysql_close($server);
它的正常工作,到目前为止,但我想知道如果我使用的最好的方法,以实现那个?
这里是jQuery脚本处理来自我的代码生成的JSON数据上面......它填充的形式与不同值的多个列表:
$.getJSON('test.php', function(result) {
var optionsValues = '';
$.each(result['mylist1'], function(item) {
optionsValues += '<option value="' + this.value + '">' + this.label + '</option>';
});
var options = $('#DestAirportList');
options.append(optionsValues);
var optionsValues = '';
$.each(result['mylist2'], function(item) {
optionsValues += '<option value="' + this.value + '">' + this.label + '</option>';
});
var options = $('#DestAirportRoomList');
options.append(optionsValues);
});
谢谢您的回答!
请格式化你的代码 – JochenJung 2010-08-26 14:08:32