也许它是重复的，但我找不到解决方案，所以我发布了这个问题。我使用 jquery ui 来自动完成搜索框。它工作正常，但问题是我想使用 id.example 进行搜索：当用户键入 paris 时，我尝试在 mysql 中发送 city_id 进行搜索。所以问题是如何用json传递隐藏的id？
这里的代码：
<input type="text" name="grno" id="grno" class="input" title="<?php echo $lng['vldgrno'];?>
jquery code:
<script>
$(function() {
function split( val ) {
return val.split( /,\s*/ );
}
function extractLast( term ) {
return split( term ).pop();
}
$( "#grno" )
// don't navigate away from the field on tab when selecting an item
.bind( "keydown", function( event ) {
if ( event.keyCode === $.ui.keyCode.TAB &&
$( this ).data( "ui-autocomplete" ).menu.active ) {
event.preventDefault();
}
})
.autocomplete({
source: function( request, response ) {
$.getJSON( "pages/search.php", {
term: extractLast( request.term )
}, response );
},
search: function() {
// custom minLength
var term = extractLast( this.value );
if ( term.length < 1 ) {
return false;
}
},
focus: function() {
// prevent value inserted on focus
return false;
},
select: function( event, ui ) {
var terms = split( this.value );
// remove the current input
terms.pop();
// add the selected item
terms.push( ui.item.value );
// add placeholder to get the comma-and-space at the end
terms.push( "" );
this.value = terms.join( "," );
alert(data.id);
return false;
}
});
});
</script>
autocomplete.php :
<?php
mysql_connect('localhost','root','');
mysql_select_db('school');
$return_arr = array();
$term = trim(strip_tags($_GET['term']));//retrieve the search term that autocomplete sends
//create select query
$query = "select `grno`,`first_name`,`student_id` from `tbl_student` where `grno` like '%".$term."%' or `first_name` like '%".$term."%'";
// Run the query
$result = mysql_query ($query);
$array = array();
while($obj = mysql_fetch_array($result)){
$array['name'] = $obj['first_name']."(".$obj['grno'].")";
array_push($return_arr,$obj['first_name']."(".$obj['grno'].")");
}
$json = json_encode($return_arr);
echo $json;
?>
so.如何传入student_id，autocomplete.php比如标签和值。{label='xyz' value='1'}