我知道以前有人问过这个问题,但我找不到任何最新或实用的答案(至少对于我的应用程序而言)。
我的 JQuery 自动完成框使用 mysql 数据库作为其源。我希望用户能够输入以获取建议,但随后被迫从下拉选项中进行选择,然后才能提交表单。
我的Javascript:
<script type="text/javascript">
$.widget( 'ui.autocomplete', $.ui.autocomplete, {
_renderMenu: function( ul, items ) {
var that = this;
$.ui.autocomplete.currentItems = items;
$.each( items, function( index, item ) {
that._renderItemData( ul, item );
});
}
});
$.ui.autocomplete.currentItems = [];
$(function() {
$("#college").autocomplete({
source: "search.php",
minLength: 5
});
});
var inputs = {college: false};
$('#college').change(function(){
var id = this.id;
inputs[id] = false;
var length = $.ui.autocomplete.currentItems.length;
for(var i=0; i<length; i++){
if($(this).val() == $.ui.autocomplete.currentItems[i].value){
inputs[id] = true;
}
}
});
$('#submit').click(function(){
for(input in inputs){
if(inputs.hasOwnProperty(input) && inputs[input] == false){
alert('incorrect');
return false;
}
}
alert('correct');
$('#college_select_form').submit();
});
</script>
我的表格:
<form action="choose.php" method="post" id="college_select_form" name="college_select_form">
<input type="text" id="college" name="college" class="entry_field" value="Type your school" onclick="this.value='';" onfocus="this.select()" onblur="this.value=!this.value?'Type your school':this.value;" /><input type="submit" id="submit" name="submit" class="submitButton" value="Go" title="Click to select school" />
</form>
搜索.php:
<?php
try {
$conn = new PDO("mysql:host=$dbhost;dbname=$dbname", $dbuser, $dbpass);
}
catch(PDOException $e) {
echo $e->getMessage();
}
$return_arr = array();
if ($conn)
{
$ac_term = "%".$_GET['term']."%";
$query = "SELECT * FROM college_list where name like :term";
$result = $conn->prepare($query);
$result->bindValue(":term",$ac_term);
$result->execute();
/* Retrieve and store in array the results of the query.*/
while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
array_push($return_arr, array('label' => $row['name'], 'value' => $row['name']));
}
}
/* Free connection resources. */
//$conn = null;
/* Toss back results as json encoded array. */
echo json_encode($return_arr);
?>
那么这样做的最佳方法是什么?我能想到的唯一解决方案是使用 PHP 来验证文本框的值是否与数据库中的值匹配,但我不确定如何用我当前的代码实现它。