1 回答

TA貢獻1830條經驗 獲得超3個贊
我認為你只是有一些語法錯誤。
<?php
session_start();
include 'sqlconnection.php';
$conn = OpenCon();
$stmt = $conn->prepare('SELECT uid,coverphoto,profilepic,bio FROM profile WHERE uid = ?');
$stmt->bind_param('i', $_SESSION['uid']);
$result = $stmt->execute();
if ($result){
if ($result['coverphoto']){
$coverphoto = 'coverphoto.jpg';
}else{
if ($result['profilepic']){
$profilepic = "profilepic.jpg";
}else{
$stmt->bind_result($uid,$coverphoto,$profilepic,$bio);
while($stmt->fetch()) {
$output=array(
'uid' => $uid,
'coverphoto' => $coverphoto,
'profilepic' => $profilepic,
'bio' => $bio
);
}
}
}
}
$json=json_encode($output);
echo $json;
$stmt->close();
CloseCon($conn);
?>
你也可以這樣做
<?php
<?php
session_start();
include 'sqlconnection.php';
$conn = OpenCon();
$stmt = $conn->prepare('SELECT uid,coverphoto,profilepic,bio FROM profile WHERE uid = ?');
$stmt->bind_param('i', $_SESSION['uid']);
$stmt->execute();
$stmt->bind_result($uid,$coverphoto,$profilepic,$bio);
while($stmt->fetch()) {
$output=array(
'uid' => $uid,
'coverphoto' => $coverphoto ? $coverphio : 'coverphoto.jpg',
'profilepic' => $profilepic ? $profilepic : 'profilepic.jpg',
'bio' => $bio
);
}
$json=json_encode($output);
echo $json;
$stmt->close();
CloseCon($conn);
?>
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