誰能幫我嗎？我想顯示完整的 mysql 數據庫表結果，但出現錯誤：我們有聯系。警告：mysqli_fetch_row() 期望參數 1 為 mysqli_result，第 28 行 C:\xampp\htdocs\test\fetch.php 中給出的字符串 $connection = mysqli_connect('127.0.0.1:3307','root','','loginapp'); if($connection){ echo "We are connected."; } $query = "SELECT * FROM users"; mysqli_query($connection,$query);?><!DOCTYPE html><html><head> <meta charset="UTF-8"> <title>Display</title> <link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/3.4.1/css/bootstrap.min.css" integrity="sha384-HSMxcRTRxnN+Bdg0JdbxYKrThecOKuH5zCYotlSAcp1+c8xmyTe9GYg1l9a69psu" crossorigin="anonymous"></head><body> <div class="container"> <div class="col-md-6"><?php while($result = mysqli_fetch_row($query)){ print_r($result); } ?> </div> </div></body></html> ```
1 回答
慕絲7291255
TA貢獻1859條經驗 獲得超6個贊
您直接在 mysqli_fetch_row() 函數中傳遞查詢，您應該傳遞 mysql 查詢
mysqli_query($connection,$query); 到 $querynew = mysqli_query($connection,$query); 并在 while($result = mysqli_fetch_row($querynew)){
像下面這樣
$connection = mysqli_connect('127.0.0.1:3307','root','','loginapp');
if($connection){
echo "We are connected.";
}
$query = "SELECT * FROM users";
$querynew = mysqli_query($connection,$query);
while($result = mysqli_fetch_row($querynew)){
print_r($result);
}
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