現在我有一個查詢返回結果列表,並將它們顯示爲使用AJAX的頁面上的鏈接。我有另一個工作查詢,我想用來比較第一個,但我不確定如何做到這一點(新的AJAX)。AJAX/XML/PHP - 比較兩個查詢的結果並顯示它們
我想要做的最終是在兩個查詢的結果中找到匹配,並格式化與不匹配風格的鏈接($(「#faramcompleted」)。append)。
PHP(xml2.php):
$query = "SELECT Name FROM judges LEFT JOIN $court
ON ($court.JudgeID = judges.JudgeID)
where Month='$month' and Year='$year' order by Name asc;";
$resultID = mysql_query($query, $linkID) or die("Data not found.");
$xml_output = "<?xml version=\"1.0\"?>\n";
$xml_output .= "<entries>\n";
for($x = 0 ; $x < mysql_num_rows($resultID) ; $x++){
$row = mysql_fetch_assoc($resultID);
$xml_output .= "\t<entry>\n";
$xml_output .= "\t\t<name>" . $row['Name']. "</name>\n";
$xml_output .= "\t</entry>\n";
}
$xml_output .= "</entries>";
echo $xml_output;
AJAX/JS:
$.ajax({
type: "POST",
url: "xml2.php",
data: 'court='+x,
dataType: "xml",
success: parseXml
});
function parseXml(xml)
{
$(xml).find("entry").each(function()
{
$("#judgesCompleted").append('<a href="viewreport.php">'+$(this).find("name").text()+'</a><br />');
});
}
我的新查詢:
$query2 = "SELECT Name FROM judges
LEFT JOIN $court ON ($court.JudgeID = judges.judgeID)
LEFT JOIN users ON ($court.userID = users.userID) WHERE Month='10' AND Year='2011' AND users.type = 'user' ORDER BY Name ASC; "
希望這是有道理的。