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我正在創建將顯示所有資產列表的資產數據庫系統（只是一個離線系統，未連接到互聯網）。在列表中，我可以點擊任何資產查看詳細信息。另外我還設法更新細節或刪除資產。但是當它轉到資產記錄部件時，它會在使用表單將記錄插入資產時發生錯誤。使用php表格在sql中插入數據時出錯
這是我的記錄添加表單。我也想讓機器ID在機器ID字段的形式下可見，但我還沒有知道如何把數據放在那裏。
插入記錄，其將捕獲的機器ID上rekod_add.php（記錄添加）地址從資產表地址突入rekod_tab表。
這裏是我記錄添加頁面（rekod_add.php）
<?php
//Start session
session_start();
//Check whether the session variable SESS_MEMBER_ID is present or not
if(!isset($_SESSION['username']) || (trim($_SESSION['password']) == '')) {
header("location: login.php");
exit();
}
?>
<html>
<head>
<title>EXA_mySQL</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<style type="text/css">
body,td,th {
font-family: Tahoma, Geneva, sans-serif;
}
</style>
</head>
<body>
<script type="text/javascript">function checkinput() {
var id_mesin = document.getElementById('id_mesin').value;
if(!id_mesin.match(/\S/)) {
alert ('Please enter Machine ID');
return false;
} else {
return true;
}
}
</script>
<?php
$con=mysqli_connect("localhost","root","admin","exa");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$id_mesin = $_POST['id_mesin'];
$query = "SELECT * FROM asset WHERE id_mesin ='".$id_mesin."'";
$result = mysqli_query($con,$query);
$rows = mysqli_fetch_array($result);
?>
<table width="733" border="0" align="center" cellpadding="0" cellspacing="1">
<tr>
<td><form name="form_insert" method="post" action="rekod_add_ac.php">
<table width="709" border="0" align="center">
<tr>
<th width="23" scope="col">MACHINE ID</th>
<th colspan="2" scope="col">DATE</th>
<th width="68" scope="col">TIME</th>
<th width="175" scope="col">RECEIVE CALL BY</th>
<th width="97" scope="col">CURRENT METER</th>
<th width="90" scope="col">LAST METER</th>
<th width="136" scope="col">J.SHEET NO</th>
</tr>
<tr>
<td> </td>
<td colspan="2"><input name="tarikh_rekod" type="text" id="tarikh_rekod" size="15" /></td>
<td><input name="time" type="text" id="time" size="10" maxlength="9" /></td>
<td><input type="text" name="call_by" id="call_by" /></td>
<td><input name="meter_semasa" type="text" id="meter_semasa" size="15" /></td>
<td><input name="meter_last" type="text" id="meter_last" size="15" /></td>
<td><input name="rujukan" type="text" id="rujukan" size="10" /></td>
</tr>
<tr>
<td> </td>
<th width="81">PROBLEM</th>
<th width="5">:</th>
<td colspan="3"><textarea name="masalah" id="masalah" cols="55" rows="5"></textarea></td>
<th colspan="2" rowspan="2"><p>REMARK</p>
<p>
<textarea name="remark" cols="30" rows="6" id="remark"></textarea>
</p></th>
</tr>
<tr>
<td> </td>
<th>SOLUTION</th>
<th>:</th>
<td colspan="3"><textarea name="solution" id="solution" cols="55" rows="5"></textarea></td>
</tr>
<tr>
<td colspan="8" align="right"><?php echo "<input type='hidden' value='" . $rows['id_mesin'] . "' name='id_mesin'>"; echo "<input type='submit' value='Add Record'>";?></td>
</tr>
</table>
</form>
</td>
</tr>
</table>
<?php
mysqli_close($con);
?>
</body>
</html>
這裏是我的rekod_add_ac.php
<?php
session_start();
if(!isset($_SESSION['username']) || (trim($_SESSION['password']) == '')) {
header("location: login.php");
exit();
}
?>
<html>
<head>
<title>EXA_mySQL</title>
<script type="text/javascript">
<!--
function CloseWindow() {
window.close();
window.opener.location.reload();
}
//-->
</script>
</head>
<body>
<?php
error_reporting(E_ALL);
ini_set('display_errors','on');
$con=mysqli_connect("localhost","root","admin","exa");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
print_r($_POST);
$id_mesin=$_POST['id_mesin'];
$tarikh_rekod=$_POST['tarikh_rekod'];
$time=$_POST['time'];
$call_by=$_POST['call_by'];
$meter_semasa=$_POST['meter_semasa'];
$meter_last=$_POST['meter_last'];
$rujukan=$_POST['rujukan'];
$masalah=$_POST['masalah'];
$solution=$_POST['solution'];
$remark=$_POST['remark'];
$rekod_in="INSERT INTO rekod_tab (id_mesin, tarikh_rekod, time, call_by, meter_semasa, meter_last, rujukan, masalah, solution, remark) VALUES ($'id_mesin', $'tarikh_rekod', $'time', $'call_by', $'meter_semasa', $'meter_last', $'rujukan', $'masalah', $'solution', $'remark')";
$result=mysqli_query($con, $rekod_in);
if($result){
echo "Successful";
echo "<BR>";
echo "<th><form>";
echo "<input type='button' onClick='CloseWindow()' value='Back to Exa_mySQL' align='middle'>";
echo "</form></th>";
}
else {
echo "Data error, please recheck before submit.";
echo "<BR>";
echo "Click back to add record.";
echo "<BR>";
echo "<form action='rekod_add.php?id=$id_mesin' method='post'>";
echo "<td><input type='hidden' value='$id_mesin' name='id_mesin'>";
echo "<input type='submit' value='Back'></td>";
echo "</form>";
echo "<th><form>";
}
mysqli_close($con);
?>
</body>
</html>
用戶完成插入記錄的詳細信息後，表單將在rekod_tab表中添加記錄，包括機器ID（id_mesin），它可以像我之前說的那樣從url自動捕獲。
但結果是錯誤的。當在sql中插入詳細的手冊時，它的工作。誰能幫我？
這裏我的錯誤結果。
對不起我的英文不好。
所有你的價值觀'$'id_mesin''等嘗試INSERT查詢有變量外的'$'跡象。 –
什麼是'$'id_mesin''？從來沒有見過任何人使用這個...應該是''$ id_mesin'' – Darren
只是通知我把'''標誌放在錯誤的地方。已經改爲''$ id_mesin'' – newbie