我想将“具有多个值的Jquery UI自动完成”应用于一个注册表单输入字段。
我想要做的事情:当访问者为此输入字段键入现有用户的名称时,首先,脚本搜索名称存在,完成它(如果存在),添加逗号。用户可以在此字段中键入第二个,第三个...现有用户名,并且每次脚本都将自动完成。当访问者点击提交按钮时,PHP搜索此用户名的id,创建id的数组,将其添加到db表中的新用户“friends”字段。
我的代码:
HTML
<form action="index.php" method="post">
<input class="std" type="text" name="friends" id="friends"/>
<input type="submit" name="submit">
</form>
Jquery的
$(function() {
function split( val ) {
return val.split( /,\s*/ );
}
function extractLast( term ) {
return split( term ).pop();
}
$( "#friends" )
// 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( "autocomplete" ).menu.active ) {
event.preventDefault();
}
})
.autocomplete({
source: function( request, response ) {
$.getJSON( "search.php", {
term: extractLast( request.term )
}, response );
},
search: function() {
// custom minLength
var term = extractLast( this.value );
if ( term.length < 2 ) {
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( ", " );
return false;
}
});
});
这是来自示例文件夹的原始php文件,完美运行。但我想从数据库而不是数组中获取 原始search.php
$q = strtolower($_GET["term"]);
if (!$q) return;
$items = array(
"Great Bittern"=>"Botaurus stellaris",
"Little Grebe"=>"Tachybaptus ruficollis",
"Black-necked Grebe"=>"Podiceps nigricollis",
"Little Bittern"=>"Ixobrychus minutus",
"Black-crowned Night Heron"=>"Nycticorax nycticorax",
"Purple Heron"=>"Ardea purpurea",
"White Stork"=>"Ciconia ciconia",
"Spoonbill"=>"Platalea leucorodia",
"Red-crested Pochard"=>"Netta rufina",
"Common Eider"=>"Somateria mollissima",
"Red Kite"=>"Milvus milvus",
);
function array_to_json( $array ){
if( !is_array( $array ) ){
return false;
}
$associative = count( array_diff( array_keys($array), array_keys( array_keys( $array )) ));
if( $associative ){
$construct = array();
foreach( $array as $key => $value ){
// We first copy each key/value pair into a staging array,
// formatting each key and value properly as we go.
// Format the key:
if( is_numeric($key) ){
$key = "key_$key";
}
$key = "\"".addslashes($key)."\"";
// Format the value:
if( is_array( $value )){
$value = array_to_json( $value );
} else if( !is_numeric( $value ) || is_string( $value ) ){
$value = "\"".addslashes($value)."\"";
}
// Add to staging array:
$construct[] = "$key: $value";
}
// Then we collapse the staging array into the JSON form:
$result = "{ " . implode( ", ", $construct ) . " }";
} else { // If the array is a vector (not associative):
$construct = array();
foreach( $array as $value ){
// Format the value:
if( is_array( $value )){
$value = array_to_json( $value );
} else if( !is_numeric( $value ) || is_string( $value ) ){
$value = "'".addslashes($value)."'";
}
// Add to staging array:
$construct[] = $value;
}
// Then we collapse the staging array into the JSON form:
$result = "[ " . implode( ", ", $construct ) . " ]";
}
return $result;
}
$result = array();
foreach ($items as $key=>$value) {
if (strpos(strtolower($key), $q) !== false) {
array_push($result, array("id"=>$value, "label"=>$key, "value" => strip_tags($key)));
}
if (count($result) > 11)
break;
}
echo array_to_json($result);
更改了search.php
$conn = mysql_connect("localhost", "user", "pass");
mysql_select_db("db", $conn);
$q = strtolower($_GET["term"]);
$query = mysql_query("select fullname from usr_table where fullname like %$q%");
$results = array();
while ($row = mysql_fetch_array($query)) {
array_push($results, $row);
}
echo json_encode($results);
自动填充对我不起作用。找不到错误的代码部分。请帮忙解决这个问题。抱歉我的英文不好
答案 0 :(得分:0)
您提到要创建一个ID数组(很可能与每个搜索过的名称相关联)?
在您的SQL语句中,包含您的ID字段。
$query = mysql_query("select id, fullname from usr_table where fullname like %$q%");
然后,而不是
array_push($results,$row)
尝试:
$results[] = array( $row[0] => $row[1] );
这应该遵循与search.php中的静态数组相同的数组结构,然后在进行JSON编码后对其进行修改。我不确定您的自动填充功能会作为结果输出哪个,因此请尝试此操作(switch row[0] with row[1]
)
$results[] = array( $row[1] => $row[0] );
编辑:
现在,我知道JSON编码的数组需要看起来像什么。这样做:
$results[] = array ( "id" => $row[0] , "label" => $row[1], "value" => $row[1] );
上面将数组附加到结果数组。让我知道这个是否奏效。它相当于array_push
:
array_push($results, array ( "id" => $row[0] , "label" => $row[1], "value" => $row[1] ));