我想將兩個變量傳遞給我的.php頁面...下拉變量工作良好。但是當我添加一個額外的變量時,它只發送一個0而不是我在表單中輸入的內容。將多個變量從HTML傳遞給PHP
我覺得我很接近這個解決方案......當我在這一行取代了許多:
xmlhttp.open("GET","getdeposit.php?q="+str + "&r=" +restaurant, true);
代替+餐廳,我用+ 101 ......那麼這頁面正確傳遞101 ......但是當我只是在我的表單中輸入的數字,則返回0
<html>
<head>
<script>
function showUser(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
}
}
xmlhttp.open("GET","getdeposit.php?q="+str + "&r=" +restaurant, true);
xmlhttp.send();
}
}
</script>
</head>
<body>
<form>
<p>
<label for="restaurant">Restaurant:</label>
<input type="text" id="restaurant" name="restaurant" placeholder="Enter Restaurant" />
</p>
<select name="users" onchange="showUser(this.value)">
<option value="">select a date</optoin>
<option value=" ">Today</option>
<option value="1">Yesterday</option>
<option value="2">Two Days Ago</option>
</select>
</form>
<div id="txtHint"><b>Select Business Date</b></div>
</body>
</html>
您是否在您的showUser函數中設置了變量餐廳?我看不到它,也許這就是問題所在。 – 2015-02-10 19:20:24