[英]PHP Mysqli no errors, no querys
我正在嘗試使用mysqli而不是mysql查詢,但是它不起作用。 Mysqli的:
$mysqli->connect($db1['host'], $db1['user'], $db1['password'], $db1['database']);
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
沒有錯誤。 如果我嘗試以下查詢:
if(isset($_POST['username']))
{
$password = $_POST['p'];
$random_salt = hash('sha512', uniqid(mt_rand(1, mt_getrandmax()), true));
$password = hash('sha512', $password.$random_salt);
if ($insert_stmt = $mysqli->prepare("INSERT INTO members (username, email, password, salt) VALUES (?, ?, ?, ?)")) {
$insert_stmt->bind_param('ssss', $username, $email, $password, $random_salt);
$insert_stmt->execute();
}
echo "Success";
}
什么都沒有插入,mysqli錯誤沒有錯誤。 表結構正確,說明成功。 我是mysqli的新手,我已經習慣了mysql。 錯誤報告有什么我想念的嗎?
最好嘗試一下,它來自php manual
<?php
$mysqli = new mysqli("localhost", "user", "password", "database");
if ($mysqli->connect_errno)
{
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli- >connect_error;
}
你必須這樣做
$password = hash('sha512', $password.$random_salt);
$insert_stmt = $mysqli->prepare("INSERT INTO members (username, email, password, salt) VALUES (?, ?, ?, ?)");
$insert_stmt->bind_param('ssss', $username, $email, $password, $random_salt);
if($insert_stmt->execute())
{
echo "Success";
}
實際上,您首先要檢查查詢,然后綁定參數,因為它只是顯示Success
。
您可以執行$ stmt-> execute(); 在這樣的if循環中:
if ($stmt->execute()){
$result = $stmt->affected_rows;
if ($result) { echo "yay" } else { echo "boo"; }
}
else {
printf("Execute error: %s", $stmt->error);
}
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