我正在 php 和 jquery 項目中工作，我使用 ajax 向服務器端發出請求現在我的 php 文件的 ajax 響應有問題我有以下 php 代碼<?phprequire 'db.php';$code = $_POST['code'];$status = 0;$count = 0;$name = "";$guarantors = 0; $sql = "SELECT * FROM clients WHERE code='$code'";$result = mysqli_query($link, $sql);if(mysqli_num_rows($result) > 0){ $row = mysqli_fetch_array($result); $name = $row['name']; $status = 1; $sqli = "SELECT * FROM guarantors WHERE client_code='$code'"; $resulti = mysqli_query($link, $sqli); if(mysqli_num_rows($resulti) > 0){ $guarantors = '['; while($rowi = mysqli_fetch_array($resulti)){ $count ++; $guarantors .= '{id:'.$rowi['id'].', name:'.$rowi['guarantor_name'].'},'; } $guarantors .= ']'; }else{ $guarantors = "0"; }}else{ $status = 0; $returnText = "??? ???????? ?? ???????? ?? ??? !";}echo json_encode(array("status"=>$status,"name"=>$name,"guarantors"=>$guarantors));?>我有這個 jquery 代碼 $.ajax({ url:"read_client.php", method:"POST", data: {code:$('.code').val()}, dataType: 'json', success:function(response){ if(response.status == 1){var result = $.parseJSON(response);console.log(result.guarantors.name) YAFloader.close() }else{ Swal.fire( 'error' ) YAFloader.close() } } });現在我想獲取每個id和namefrom guarantors，但我做不到我需要幫助，請快點，有人可以幫助我嗎？？？謝謝你們
1 回答
陪伴而非守候
TA貢獻1757條經驗 獲得超8個贊
$guarantors應該是一個數組：
$guarantors = [];
if (mysqli_num_rows($resulti) > 0) {
while ($rowi = mysqli_fetch_array($resulti)) {
$count++;
$guarantors[] = [
'id' => $rowi['id'],
'name' => $rowi['guarantor_name'],
];
}
}
然后在js端：
result.guarantors.map(guarantor => {
console.log(guarantor.id, guarantor.name);
})
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