-1
我已經多次試過讓這段代碼運行並將數據插入到我的數據庫中。無論我嘗試什麼,我都無法弄清楚問題所在。 PHP的是這樣的:插入表單數據到mysql數據庫
<?php
// Create connection
$conn = mysqli_connect("localhost","nmhsmusi_admin" , "********", "nmhsmusi_musicdb");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if (isset($_POST['submit']))
{
$titleTag = $_POST['title'];
$composerTag = $_POST['composer'];
$unicodeTag = $_POST['unicode'];
$tempoTag = $_POST['tempo'];
$yearTag = $_POST['year-used'];
$languageTag = $_POST['language'];
$keyTag = $_POST['key-signature'];
$pianoTag = $_POST['piano'];
$temposelTag = $_POST['temposel'];
$partsTag = $_POST['parts'];
$run = mysqli_query($conn,"INSERT INTO musicdb (title, composer, unicode, temptxt, yearused, languages, pianokeys, piano, temposel, parts)
VALUES
(
'$titleTag', '$composerTag', '$unicodeTag', '$tempoTag', '$yearTag', '$languageTag', '$keyTag', '$pianoTag', '$temposelTag', '$partsTag'
)");
if ($run) {
echo "New record created successfully";
} else {
echo "failed";
}
mysqli_close($conn);
}
?>
任何幫助,將不勝感激
你有幾個方法來檢查錯誤,你需要使用它們。該表單遺漏了這個順便說一句。 –