Php 在mysql中上载图像时给出空值
无法上载。上载图像时出现获取错误Php 在mysql中上载图像时给出空值,php,mysql,database,image,Php,Mysql,Database,Image,无法上载。上载图像时出现获取错误 echo "That's not an image."; else { if (!$insert = mysqli_query($db ,"INSERT INTO uploadimages (`Name`,`image`,`user_id` ) VALUES ('$image_name','$image',$user_id)")) echo "Problem uploading image."; else ==================
echo "That's not an image.";
else
{
if (!$insert = mysqli_query($db ,"INSERT INTO uploadimages (`Name`,`image`,`user_id` ) VALUES ('$image_name','$image',$user_id)"))
echo "Problem uploading image.";
else
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http://jsfiddle.net/amibhop/ja9fpo5b/
<form id="form" action="" method="post" enctype="multipart/form-data" name="a">
<table width="65%" border="1" align="center" cellpadding="0" cellspacing="0" bordercolor="#EFEFEF">
<input type= "file" name="image"><p>
<td height="243" colspan=3 valign="top">
<table width="100%" border=0 cellpadding="3">
<tr>
<td width="100%" align="left" class="desc"><span class="style5">
<input type="button" name="Add" value="Add" onclick="addRow();"/>
<input name="submit" type="button" id="submit" value="Delete" onclick="deleteRow1('tblSample1');" />
</span></td>
</tr>
<tr>
<td align="left" class="desc"> <table id = "tblSample1" >
</table>
<input type="hidden" name="tblid" /><input type="hidden" name="cid" value="<?php echo $_REQUEST['id']; ?>" /></td>
</tr>
<tr bgcolor="white">
<td height="28" colspan="3" align="center"><div align="center">
<input type="submit" name="submit2" value="Save" class="button1" />
</div></td>
</tr>
</table></td></tr>
</table>
</form>
</body>
<?php
//connect to database
include('1.php');
include("db.php");
if(isset($_POST['submit2']))
{
$user_id=$_POST['cid'];
// file properties
$file = $_FILES["image"]["tmp_name"];
if (!isset($file))
echo "Please select an image.";
else
{
$image =addslashes(file_get_contents($_FILES["image"]["tmp_name"]));
$image_name = addslashes($_FILES["image"]["name"]);
$image_size = getimagesize($_FILES["image"]["tmp_name"]);
if ($image_size==FALSE)
echo "That's not an image.";
else
{
if (!$insert = mysqli_query($db ,"INSERT INTO uploadimages (`Name`,`image`,`user_id` ) VALUES ('".$image_name."','".$image."','".$user_id.")"))
echo "Problem uploading image.";
else
{
$lastID= mysqli_query($db ,'select * from uploadimages ORDER BY ID DESC LIMIT 1');
$las=mysqli_fetch_array($lastID) ;
$rea=$las['ID'];
echo "Image uploaded <p> Your image:</p>echo <img src='uploadphoto1.php'?ID='$rea'>";
}
}
}
}
?>
</html>
http://jsfiddle.net/amibhop/ja9fpo5b/
由于代码不完整,因此我将向您介绍该过程
收到请求后,获取文件名并将其存储在变量中。
将文件移动到文件夹中,然后将图像路径存储在数据库中
如果您遇到mysql错误,那么可能会出现多个问题。
1.表中有一个必填字段,您没有输入任何值。
2.列名不匹配或表名错误。
3.数据库连接也可能是一个问题
请使用mysqli\u error($db)
查看问题的原因。首先指出错误,然后再解决它
您面临的问题可以通过这种方法解决
$imgname=$_FILES['image']['name'];
if($_FILES['image']['error']==0)
{
$uploadFile=move_uploaded_file($_FILES['image']['tmp_name'],"/images/$imgname");
if($uploadFile)
{
if (!$insert = mysqli_query($db ,"INSERT INTO uploadimages
(`Name`,`image`,`user_id` ) VALUES
('".$imgname."','".$imgname."','".$user_id.")"))
echo "Problem uploading image.";
else
{
$lastID= mysqli_query($db ,'select * from uploadimages ORDER BY ID DESC LIMIT 1');
$las=mysqli_fetch_array($lastID) ;
$rea=$las['ID'];
echo "Image uploaded <p> Your image:</p>echo <img src='uploadphoto1.php'?ID='$rea'>";
}
}
$imgname=$\u文件['image']['name'];
如果($\u文件['image']['error']==0)
{
$uploadFile=移动上传的文件($UPLOADFILES['image']['tmp\u name'],“/images/$imgname”);
如果($uploadFile)
{
如果(!$insert=mysqli\u查询($db),则插入到上传的图像中
(`Name`、`image`、`user_id`)值
(“$imgname.”、“$imgname.”、“$user_id.”)等)
echo“上传图像时出现问题。”;
其他的
{
$lastID=mysqli_query($db,'select*from uploadimages ORDER BY ID DESC LIMIT 1');
$las=mysqli\u fetch\u数组($lastID);
$rea=$las['ID'];
echo“上传图像您的图像:echo”;
}
}
我提供的链接可能重复您提供的链接只有html和css。您面临的是一个数据库和PHP问题。编辑您的问题并添加代码的相关部分以获得更好的帮助。您可以在js中看到这些内容。您可以想说多少次就说多少次,但我没有看到任何与PHP或sql相关的内容在链接中。如果可以,请在此处添加代码,或者尝试说服所有人在此处看到不可见的代码。我已经根据您的输入为您提供了解决方案。chk pl添加了几行代码