我想像這樣動態創建一個 JSON：{"storie":[{"story_id":"111","username":"Username1","profile_photo":"boh.png","copertina":"ok.png","num_elements":"1","rating":"4.5"},{"story_id":"222","username":"Username2","profile_photo":"hello.png","copertina":"hi.png","num_elements":"2","rating":"3.5"}]}我正在嘗試從 MySQL 數據庫獲取值，并且我能夠做到這一點：$response = array();$sql = mysqli_query($conn, "SELECT * FROM storie WHERE userid IN (SELECT following FROM follow WHERE follower='$userid')");while($row = mysqli_fetch_assoc($sql)){ $usern = getuserinfo($row['userid'], "username", $conn); $prof_photo = getuserinfo($row['userid'], "profile_photo", $conn); $idsto=$row['storia_id']; $elem = mysqli_query($conn, "SELECT COUNT(*) AS da_vedere FROM `storie_images` WHERE storia_id='$idsto' AND imm_id NOT IN (SELECT imm_id FROM image_views WHERE viewer='$userid')"); while($ok = mysqli_fetch_assoc($elem)){ $num_elem = $ok['da_vedere']; } //here I put the line that add the array to the json array:}但問題是這一行，它應該創建一個新數組并將其放入 json 中：$response['storie'] = [array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5")];當只有一條記錄時它有效，但如果有更多記錄它就不起作用。有人能幫我嗎？
2 回答
拉風的咖菲貓
TA貢獻1995條經驗 獲得超2個贊
只需替換：
$response['storie'] = [array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5")];和
$response['storie'][] = array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5");
呼啦一陣風
TA貢獻1802條經驗 獲得超6個贊
那是因為您在 while 內使用 $response['story'] 來解決此問題：
1-創建另一個名為 result 的數組，然后在 while 中使用它
$response = array();
$result = array ()
while (....)
{
//here I put the line that add the array to the json array:
$result[] = array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5");
}
2-然后在循環外使用響應數組：
$response['storie'] = $result;
你的代碼將是這樣的：
$response = array();
$result = array ()
$sql = mysqli_query($conn, "SELECT * FROM storie WHERE userid IN (SELECT following FROM follow WHERE follower='$userid')");
while($row = mysqli_fetch_assoc($sql)){
$usern = getuserinfo($row['userid'], "username", $conn);
$prof_photo = getuserinfo($row['userid'], "profile_photo", $conn);
$idsto=$row['storia_id'];
$elem = mysqli_query($conn, "SELECT COUNT(*) AS da_vedere FROM `storie_images` WHERE storia_id='$idsto' AND imm_id NOT IN (SELECT imm_id FROM image_views WHERE viewer='$userid')");
while($ok = mysqli_fetch_assoc($elem)){
$num_elem = $ok['da_vedere'];
}
//here I put the line that add the array to the json array:
$result[] = array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5");
}
$response['storie'] = $result;
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