我正在嘗試將值存儲到我的數據庫中。這是我用來從我的網站插入代碼的代碼：<?php$ident = $_POST['ident'];$dato = $_POST['dato'];$kundensnavn = $_POST['kundensnavn'];$gsm = $_POST['gsm'];$fodselsdato = $_POST['fodselsdato'];$prisplan = $_POST['prisplan'];$operator = $_POST['operator'];$portering = $_POST['portering'];$epost = $_POST['epost'];? ? // Database connection? ? $conn = new mysqli('localhost','my_username','my_password','id14293554_rw2');? ? if($conn->connect_error){? ? ? ? echo "$conn->connect_error";? ? ? ? die("Connection Failed : ". $conn->connect_error);? ? } else {? ? ? ? $stmt = $conn->prepare("INSERT INTO sales_table(ident_column, date_column, name_column, gsm_column, birthdate_column, pp_column, carrier_column, transfer_column, email_column) values(?, ?, ?, ?, ?, ?, ?, ?, ?)");? ? ? ? $stmt->bind_param("sssssssss", $ident, $dato, $kundensnavn, $gsm, $fodselsdato, $prisplan, $operator, $portering, $epost);? ? ? ? $execval = $stmt->execute();? ? ? ? echo $execval;? ? ? ? echo "Done!";? ? ? ? $stmt->close();? ? ? ? $conn->close();? ? }?>如果需要的話，這是我的index.html：<!DOCTYPE html><html ><head>? <meta charset="UTF-8">? <title>Test</title>? <link rel="stylesheet" href="./style.css"></head><body><!-- partial:index.partial.html --><div class="container">??? <form id="contact" action="insert.php" method="post">? ? <h3><center>Reg</center></h3>? ??? ? <fieldset>? ? ? <input type="text" placeholder="Ident" name="ident" required autofocus>? ? </fieldset>? ??? ? <fieldset>? ? ? <input type="text" placeholder="Dato" name="dato" required>? ? </fieldset>? ??? ? <fieldset>? ? ? <input type="text" placeholder="Kundens navn" name="kundensnavn" required>? ? </fieldset>? ??當我按下“提交”按鈕時，該腳本將運行，并給出“完成！”，因此沒有錯誤。但是，當我檢查表時，即使我輸入了所有需要的信息，那里也什么也沒有：
1 回答
阿晨1998
TA貢獻2037條經驗 獲得超6個贊
您的表包含Integer列gsm_column 但您string在此處傳遞值：
$stmt->bind_param("sssssssss", $ident, $dato, $kundensnavn, $gsm, $fodselsdato, $prisplan, $operator, $portering, $epost); <---------
嘗試改成這樣：
$stmt->bind_param("sssisssss", $ident, $dato, $kundensnavn, $gsm, $fodselsdato, $prisplan, $operator, $portering, $epost);
我也會改變最后一部分
if ($stmt->execute()) {
// it worked
echo "Done";
} else {
// it didn't
echo "Not inserted";
//Print error
echo $stmt->error;
}
$stmt->close();
$conn->close();
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