[英]Not updating SQL database
由於某種原因,我的代碼沒有更新mySQL數據庫,但它沒有報告任何錯誤。
register.php (表格)
<form class="register_form" action="action.php?do=register" method="post">
Team Name*: <input type="text" name="teamname" required />
Team Region*: <input type="text" name="teamregion" maxlength="4" required />
Team Leader*: <input type="text" name="teamleader" maxlength="16" required />
Team Members: <input type="text" name="teammembers" />
<input name="register_submit" type="submit" value="Register" />
</form>
action.php的
<?php
$con=mysqli_connect("192.185.#.###","########_reg","#######","#########");
if (mysqli_connect_errno()) {echo "Failed to connect to MySQL: " . mysqli_connect_error();}
$action = $_GET['do'];
if($action=="register") {
$teamname = $_POST["teamname"];
$teamregion = $_POST["teamregion"];
$teamleader = $_POST["teamleader"];
$teammembers = $_POST["teammembers"];
$result = mysqli_query($con, "INSERT INTO teams (teamname, region, teamleader, teammembers, wins, loses)
VALUES (" . $teamname . "," . $teamregion . "," . $teamleader . "," . $teammembers . ",0,0);");
}
?>
任何想法為什么這不能正常工作?
這是一個帶有預處理語句的工作示例,通常使用“更好”而不是query
action.php的
$con = new mysqli('localhost', 'root', '', 'dachi');
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
if (isset($_GET['do']) && $_GET['do'] === "register") {
$teamname = $_POST["teamname"];
$teamregion = $_POST["teamregion"];
$teamleader = $_POST["teamleader"];
$teammembers = $_POST["teammembers"];
$wins = 0;
$loses = 0;
$stmt = $con->prepare("INSERT INTO `teams` (`teamname`,`region`,`teamleader`,`teammembers`,`wins`,`loses`) VALUES (?,?,?,?,?,?)");
$stmt->bind_param('ssssii', $teamname, $teamregion, $teamleader, $teammembers, $wins, $loses);
$stmt->execute();
$stmt->close();
}
register.php
<form class="register_form" action="action.php?do=register" method="post">
Team Name*: <input type="text" name="teamname" required />
Team Region*: <input type="text" name="teamregion" maxlength="4" required />
Team Leader*: <input type="text" name="teamleader" maxlength="16" required />
Team Members: <input type="text" name="teammembers" />
<input name="register_submit" type="submit" value="Register" />
</form>
你應該改變:
if($action=="register") {
至
if($action=="register_submit") {
因為您的輸入標記名稱設置為值register_submit not register 。
並改變$action = $_GET['do'];
to $action = $_POST['register_submit'];
register.php(表格)
<form class="register_form" action="action.php?do=register" method="post">
action.php的
$action = $_GET['do'];
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