我试图将两个不同的值传递给XMLHttpRequestObject.send
,但似乎无法让php脚本同时读取它们,它只会读取传递的第一个值。如何让 PHP 读取这两个值,下面是示例代码:
HTML
<select id="mes" onchange="callCourseYards(this);">
<option>-</option>
</select>
的Javascript
var XMLHttpRequestObject = false;
if(window.XMLHttpRequest){
XMLHttpRequestObject = new XMLHttpRequest();
}else if(window.ActiveXObject){
XMLHttpRequestObject = new ActiveXObject("Microsoft.XMLHTTP");
}
function callCourseYards(selectObject){
if(XMLHttpRequestObject){
XMLHttpRequestObject.open("POST", "php/CourseChoiceTeeOff.php");
XMLHttpRequestObject.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
XMLHttpRequestObject.onreadystatechange = function(){
if(XMLHttpRequestObject.readyState == 4 && XMLHttpRequestObject.status == 200){
var returnData = XMLHttpRequestObject.responseText;
var messageDiv = document.getElementById('tee');
messageDiv.innerHTML = returnData;
}
}
var data = selectObject.value;
var course = selectObject.value;
XMLHttpRequestObject.send("data=" + data + "course=" + course);
}
return false;
}
PHP
<?php
$pdo_dsn='localhost';
$dbname ='XXX';
$mysql_user='XXX';
$mysql_pass='XXX';
$db = mysqli_connect($pdo_dsn, $mysql_user, $mysql_pass, $dbname);
if(!$db){
print "Unable to connect to MySQL";
}
$myData = $_POST['data'];
$myData1 = $_POST['course'];
$sql_state = "SELECT * FROM ".$myData1." WHERE CourseCode = '".$myData."' ORDER BY CourseCode";
$result = mysqli_query($db, $sql_state);
$out = "";
$myrowcount = 0;
if(!$result){
$out = "Error";
}else{
$out = "<option>Selecciona un campo</option><br/ >";
$numresults = mysqli_num_rows($result);
for ($i = 0; $i < $numresults; $i++){
$row = mysqli_fetch_array($result);
$teeOff = $row['TeeOff'];
$out .= "<option id='".$myData."' value='".$teeOff."'> ".$teeOff."</option>";
}
$out .= "<br/ >";
}
print $out;
?>
我似乎无法通过XMLHttpRequestObject
获取值进入 PHP 。我不确定它的语法或它们应该如何进入 PHP 。
我尝试过使用&amp;以及用于连接两个值的+符号,但我不确定这是否可行。有谁知道如何解决这个问题?我需要两个值才能在数据库中执行 mySQL 搜索。
非常感谢任何帮助。感谢