當(dāng)用戶從 iOS 應(yīng)用程序發(fā)送登錄命令時，我試圖從數(shù)據(jù)庫中檢索信息。為了測試這個功能，我手動啟動我的 php 頁面（例如http://www.testdatabase.com/LoginFunctions.php）并以編程方式強制使用用戶名。問題是 mysqli_query 返回 NULL 值。如果我使用“or die(mysql_error()”，什么也不會發(fā)生。即使我使用 mysqli_num_rows 返回 1，但 $result 仍然是空的。所以當(dāng) mysql_fetch_assoc 被執(zhí)行時，程序會崩潰而不顯示任何錯誤。知道嗎？謝謝<?php // Create connection $con=mysqli_connect("localhost","super","super","testdb"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $action = "login"; $username = "Peperoncino"; $response = array(); if ($action == "login") { $query = "SELECT psw AS pswrd,id FROM Activities WHERE nome = 'Peperoncino' LIMIT 1"; if ($result = mysqli_query($con, $query)) { $values = mysql_fetch_assoc($result); $password = $values['pswrd']; $response["password"] = $password; $response["message"] = "Get information from db"; } else { echo "err"; } echo json_encode($response); } // Close connections mysqli_close($con);?>
1 回答
叮當(dāng)貓咪
TA貢獻1776條經(jīng)驗 獲得超12個贊
您正在使用已棄用的 mysql_fetch 函數(shù)。使用新的
<?php
// Create connection
$con=mysqli_connect("localhost","super","super","testdb");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$action = "login";
$username = "Peperoncino";
$response = array();
if ($action == "login")
{
$query = "SELECT psw AS pswrd,id FROM Activities WHERE nome = 'Peperoncino' LIMIT 1";
if ($result = mysqli_query($con, $query))
{
$values = mysqli_fetch_assoc($result);
$password = $values['pswrd'];
$response["password"] = $password;
$response["message"] = "Get information from db";
}
else
{
echo mysqli_error($conn);
}
echo json_encode($response);
}
// Close connections
mysqli_close($con);
?>
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