Question

我输入了id aktForm_tiekejas并且是jquery自动完成代码：

$('#aktForm_tiekejas').autocomplete({
serviceUrl: '_tiekejas.php',
width: 185,
deferRequestBy: 0,
noCache: true,
onSelect: function(suggestion) {alert('You selected:'+suggestion.value+','+suggestion.data);}
});

_tiekejas.php：

<?php
include("../Setup.php");
$query = ($_GET['query']);
$reply = array();
$reply['query'] = $query;
$reply['suggestions'] = array();
$reply['data'] = array();
$res = mysql_query("SELECT id,pavadinimas FROM sarasas_tiekejas WHERE pavadinimas LIKE '%$query%' ORDER BY pavadinimas ASC");
while ($row = mysql_fetch_array($res)) {
 $reply['suggestions'][] = $row['pavadinimas'];
 $reply['data'][] = $row['id'];
}
mysql_close();
echo json_encode($reply);
?>

如果查询是'vac'php从服务器返回：

{"query":"vac","suggestions":["UAB Vivacitas"],"data":["866"]}

但是

alert('You selected:'+suggestion.value+','+suggestion.data);

不会提醒数据（866）

为什么呢？......

Answer 1

可能是因为suggestion.value不存在。查看您的JSON响应代码，我可以看到suggestion.data但没有advicetion.value。由于JS将寻找不存在的值，因此会抛出错误。你也错过了返回第二部分的数组。试试这个：

alert('You selected: '+suggestion.data[0]);

如果您需要遍历数据子集，请执行以下操作：

for(i in suggestion.data){
    alert('You selected: '+suggestion.data[i]);
}

jquery autocomplete不会返回建议数据onselect

1 个答案: