我通過 jquery ajax發布PHP文件（ ）。post.php我想以 javascript 變量的形式從中獲取數據。我成功地在我的控制臺中獲取了數據。但我不知道如何使用這個變量。你可以在下面看到我的代碼。$.post( "post.php", { region: region, district: district }, function(data) { console.log(data); } );我的 post.php 頁面看起來像這樣 @include('../../_partials/_dbConnect.php'); $region = $_POST['region']; $district = $_POST['district']; $sql = "SELECT * FROM table1 WHERE name_rg= '".$region."'"; $result = pg_query($db_connection, $sql); while ($row = pg_fetch_row($result)) { $cols = array($row[0],$row[1],$row[2],$row[3],$row[4],$row[5],$row[6],$row[7],$row[8],$row[9],$row[10],$row[11],$row[12],$row[13],$row[14],$row[15],$row[16],$row[17],$row[18],$row[19],$row[20],$row[21]); }<script> var cols = [<?php echo '"'.implode('","', $cols).'"' ?>];</script>console.log(data)像這樣的輸出，<script> var cols = ["94","32","361","0","118","159","0","243","702","1775","8","0","2","0","150","135","381","2","0","0","0","0"];</script>非常感謝您的幫助。
2 回答
紅顏莎娜
TA貢獻1842條經驗 獲得超13個贊
在您的post.php中，您可以簡單地回顯數組，jQuery 應該自動將其轉換array為響應
// post.php
<?php
@include('../../_partials/_dbConnect.php');
$region = $_POST['region'];
$district = $_POST['district'];
$sql = "SELECT * FROM table1 WHERE name_rg= '".$region."'";
$result = pg_query($db_connection, $sql);
while ($row = pg_fetch_row($result)) {
$cols = array($row[0],$row[1],$row[2],$row[3],$row[4],$row[5],$row[6],$row[7],$row[8],$row[9],$row[10],$row[11],$row[12],$row[13],$row[14],$row[15],$row[16],$row[17],$row[18],$row[19],$row[20],$row[21]);
}
echo json_encode($cols);
?>
// Somewhere in your js
$.post(
"post.php",
{
region: region,
district: district
},
function(data) {
console.log(data[0]);
}
);
慕碼人8056858
TA貢獻1803條經驗 獲得超6個贊
在 javascript 中使用 JSON.parse()
$.post(
"post.php",
{
region: region,
district: district
},
function(data) {
data=JSON.parse(data);
}
);```
And here you go, u can play with it as you need
Happy learning!
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