Question

我希望在使用add_data（）函数插入数据后显示从数据库加载的数据，但它不会在表格中显示数据。

if(isset($_POST['btn_add_details'])) {
add_data();
include("db_PSIS.php");
$sql2="SELECT * FROM sample_barcode WHERE IDRec='".$row['IDRec']."'";
$result2=mysql_query($sql2);
if(!$result2) {
echo "<h1>Could not process query this time. Please try again later!</h1>";
}
else {
while($row2=mysql_fetch_array($result2)) { 

echo "<form name='form2' method='POST'>";
echo "<table class='output' border=2 align=center>";
echo "<tr class='thcolor'>";
echo "<th>Parent</th>";
echo "<th>LOT Traveller No.</th>";
echo "<th>Datecode</th>";
echo "</tr>";
echo "<tr>";
echo "<td>".$row2['LotTraveller']."</td>";
echo "<td>".$row2['LotTraveller']."</td>";
echo "<td>".$row2['Datecode']."</td>";
echo "</tr>";
echo "</table>";
echo "</form>";
}
echo "<br><h1>Data successfully loaded!</h1>";
}
mysql_close($link);
}
?>

这是在数据库中添加数据的功能：

function add_data() {

include("db_PSIS.php");
$sql="INSERT INTO sample_barcode (LotTraveller, ShipmentLotNumber) VALUES ('".$_POST['traveller']."', '".$_POST['datecode']."')";
$result=mysql_query($sql);
$row=mysql_fetch_array($result);
mysql_close($link);

}

这是数据库连接的代码：

<?php
$username="****";
$password="****";
$database="****";
$hostname="****";
$link=@mysql_connect($hostname,$username,$password)
or die ("...cannot connect database, using $username for $hostname");
mysql_query("set names 'utf8'"); //指定数据库字符集
mysql_select_db($database,$link);

＆GT;

这是表单输入：

<table width="500" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF">
<tr>

<tr>
<th align="right">Parent Traveller: </th>
<td><input type="text" name="traveller" size="20" value="<?php if(empty($_POST['traveller'])) {echo ''; } else { echo $row2['LotTraveller']; } ?>"/></td>
</tr>
<tr>
<th align="right">Date Code: </th>
<td><input type="text" name="datecode" size="20" value="<?php if(empty($_POST['traveller'])) {echo ''; } else { echo $row2['ShipmentLotNumber']; } ?>" /></td>
</tr>
<tr>
<th align="right">Traveller 1: </th>
<td><input type="text" name="traveller1" size="20" /></td>
</tr>
<tr>
<th align="right">Date Code: </th>
<td><input type="text" name="datecode1" size="20" /></td>
</tr>
<tr>
<th align="right">Traveller 2: </th>
<td><input type="text" name="traveller2" size="20" /></td>
</tr>
<tr>
<th align="right">Date Code: </th>
<td><input type="text" name="datecode2" size="20" /></td>
</tr>
<tr>
<th align="right"> </th>
<td><input type="hidden" name="id" size="20" value="<?php if(empty($_POST['traveller'])) {echo ''; } else {echo $row2['IDRec'];} ?>"/></td>
</tr>
<tr>
<td colspan="2" align="center"><input type="submit" name="btn_add_details" id="btn_add_details" onClick="return check_submit();" value="Add the Detail(s)" onfocus="setStyle(this.id)"
    style="color: #000000;
    padding: 2px 5px;
    border: 2px solid;
    border-color: #7bf #07c #07c #4AA02C;
    background-color: #09f;
    font-family: Georgia, ..., serif;
    font-size: 18px;
    display: block;
    height: 30px;
    width: 200px;" /> </td>
</tr>
</table>

Answer 1

既然您声明IDRec是您的PK并且设置为自动递增，并且您正在根据单个数字提取记录...您将永远不会在结果集中获得多个结果。

更改此代码块...

$sql2="SELECT * FROM sample_barcode WHERE IDRec='".$row['IDRec']."'";
$result2=mysql_query($sql2);
if(!$result2) {
echo "<h1>Could not process query this time. Please try again later!</h1>";
}
else {
while($row2=mysql_fetch_array($result2)) { 

echo "<form name='form2' method='POST'>";
echo "<table class='output' border=2 align=center>";
echo "<tr class='thcolor'>";
echo "<th>Parent</th>";
echo "<th>LOT Traveller No.</th>";
echo "<th>Datecode</th>";
echo "</tr>";
echo "<tr>";
echo "<td>".$row2['LotTraveller']."</td>";
echo "<td>".$row2['LotTraveller']."</td>";
echo "<td>".$row2['Datecode']."</td>";
echo "</tr>";
echo "</table>";
echo "</form>";
}
echo "<br><h1>Data successfully loaded!</h1>";
}
mysql_close($link);
}

对此...

$sql2="SELECT * FROM sample_barcode WHERE IDRec=".$row['IDRec']."";
$result2=mysql_query($sql2);
$row2=mysql_fetch_assoc($result2); // added this line
if(!$result2) {
echo "<h1>Could not process query this time. Please try again later!</h1>";
}
else {

echo "<form name='form2' method='POST'>";
echo "<table class='output' border=2 align=center>";
echo "<tr class='thcolor'>";
echo "<th>Parent</th>";
echo "<th>LOT Traveller No.</th>";
echo "<th>Datecode</th>";
echo "</tr>";
echo "<tr>";
echo "<td>".$row2['LotTraveller']."</td>";
echo "<td>".$row2['LotTraveller']."</td>";
echo "<td>".$row2['Datecode']."</td>";
echo "</tr>";
echo "</table>";
echo "</form>";
echo "<br><h1>Data successfully loaded!</h1>";
}
mysql_close($link);
}

在第一行中，我删除了IDRec周围的单引号 - 这是一个整数，因此不需要单引号。

删除了你的while循环 - 你只会得到一个结果，所以不需要循环。

此外，这似乎不是一个表单，因此您也可以删除<form> </form>代码。

最后，开始学习pdo_mysql。 my_sql已被弃用。任何仍在使用此代码的人都会在一天醒来发现他们的网站已经神奇地停止运作。

修改

在add_data函数中，将其更改为此...

function add_data() { include("db_PSIS.php"); $sql="INSERT INTO sample_barcode (LotTraveller, ShipmentLotNumber) VALUES ('".$_POST['traveller']."', '".$_POST['datecode']."')"; $result=mysql_query($sql); $sql="SELECT * FROM sample_barcode ORDER BY IDRec DESC"; // added this line $result=mysql_query($sql); // added this line $row=mysql_fetch_assoc($result); // changed this line from array to assoc mysql_close($link); }

这将获得您刚刚输入的数据的结果。我添加了一个查询来获取您需要显示的数据。

快乐的编码！

用于在表上显示数据的MYSQL Select查询无效

1 个答案: