我想使用表单将多个图像添加到 mysql 数据库中。每当用户提交该表单时，它就会捕获所有数据并将其存储到 mysql 数据库中。（这是正确的方式，但我需要）
我的数据库表看起来像这样。
CREATE TABLE `Owner_detail` (
`id` int(10) NOT NULL auto_increment,
`fullname` varchar(30) NOT NULL,
`List1` varchar(20) NOT NULL,
`List2` varchar(20) NOT NULL,
`List3` varchar(20) NOT NULL,
`area` varchar(30) NOT NULL,
`ownermobile` varchar(20) NOT NULL,
`email` varchar(20) NOT NULL,
`image_one` blob NOT NULL,
`image_two` blob NOT NULL,
`image_three` blob NOT NULL,
`otherdetail` varchar(300) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=49 ;
我的 html 表单看起来像这样。
<table width="60%"><form action="process.php" method="post" enctype="multipart/form-data" name="tripleplay" onsubmit="MM_validateForm('fullname','','R','area','','R','ownermobile','','RisNum','email','','RisEmail','otherdetail','','R');return document.MM_returnValue">
<tr>
<td width="44%">Full Name : </td>
<td width="56%"><input size="25" type="text" name="fullname" id="fullname" /></td>
</tr>
<tr>
<td valign="top">Rental Type:</td>
<td>
<select name='List1' id="List1" onchange="fillSelect(this.value,this.form['List2'])">
<option selected>Make a Selection</option>
</select>
<br /><br />
<select name='List2' id="List2" onchange="fillSelect(this.value,this.form['List3'])">
<option selected>Make a Selection</option>
</select><br /><br />
<select name='List3' id="List3" onchange="getValue(this.value, this.form['List2'].value,
this.form['List1'].value)">
<option selected>Make a Selection</option>
</select>
</td>
</tr>
<tr>
<td>Area : </td>
<td><input size="25" type="text" name="area" id="area" />
sq.ft.</td>
</tr>
<tr>
<td>Owner Mobile / Landline</td>
<td><input size="25" type="text" name="ownermobile" id="ownermobile" /></td>
</tr>
<tr>
<td><p>E-mail</p></td>
<td><input size="25" type="text" name="email" id="email" /></td>
</tr>
<tr>
<td valign="top">Pictures of Property</td>
<td>
<input type="file" name="image_one" id="image_one" />
<input type="file" name="image_two" id="image_two" />
<input type="file" name="image_three" id="image_three" />
</td>
</tr>
<tr>
<td valign="top">Other details you would like to share about your property:</td>
<td><textarea size="25" name="otherdetail" id="otherdetail" cols="45" rows="5"> </textarea></td>
</tr>
<tr>
<td> </td>
<td><input type="submit" name="submit" id="submit" value="Submit" /></td>
</tr>
</form>
</table>
我在 php 中的处理形式是这样的
<?
if( $_POST )
{
$con = mysql_connect("","","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("vkrental", $con);
$users_fullname = $_POST['fullname'];
$users_area = $_POST['area'];
$users_List1 = $_POST['List1'];
$users_List2 = $_POST['List2'];
$users_List3 = $_POST['List3'];
$users_ownermobile = $_POST['ownermobile'];
$users_email = $_POST['email'];
$users_image_one = $_POST['image_one'];
$users_image_two = $_POST['image_two'];
$users_image_three = $_POST['image_three'];
$users_otherdetail = $_POST['otherdetail'];
$users_fullname = htmlspecialchars($users_fullname);
$users_area = htmlspecialchars($users_area);
$users_List1 = htmlspecialchars($users_List1);
$users_List2 = htmlspecialchars($users_List2);
$users_List3 = htmlspecialchars($users_List3);
$users_ownermobile = htmlspecialchars($users_ownermobile);
$users_email = htmlspecialchars($users_email);
$users_image_one = htmlspecialchars($users_image_one);
$users_image_two = htmlspecialchars($users_image_two);
$users_image_three = htmlspecialchars($users_image_three);
$users_otherdetail = htmlspecialchars($users_otherdetail);
$query = "
INSERT INTO `vkrental`.`Owner_detail` (
`fullname` ,
`area` ,
`List1` ,
`List2` ,
`List3` ,
`ownermobile` ,
`email` ,
`image_one` ,
`image_two` ,
`image_three` ,
`otherdetail`
)
VALUES ( '$users_fullname',
'$users_area', '$users_List1','$users_List2','$users_List3', '$users_ownermobile', '$users_email', '$users_image_one','$users_image_two','$users_image_three', '$users_otherdetail'
);";
mysql_query($query);
echo "<h2>Success.</h2>";
mysql_close($con);
}
?>
我的问题是当我从它成功存储的 mysql 数据库中插入图像时。但是当我从我的 html 表单中执行此操作时，它不会存储图像。它只存储数据（文本）。
我不知道我的表格中的实际错误是什么。