[英]fetch column values from drop down and insert it in another table
i have created a dropdown in which i am fetching data from one table 'category'(it has two column cat_id and category name) and i want to insert column value in another table GALLERY ..i am able to fetch cat_id but can't fetch categoryname ..please help..SQL injection not a problem 我创建了一个下拉列表,其中我要从一个表“ category”(具有两列cat_id和类别名称)中获取数据,并且我想在另一张表GALLERY中插入列值。我可以获取cat_id,但不能获取类别名称..请帮助..SQL注入不是问题
<?php
include_once("header.php");
include ("connection.php");
if(isset($_REQUEST['ansave']))
{
$a=$_REQUEST['choosecategory'];
$image=$_FILES['uploadgallery']['name'];// name given in input type
$ext=substr(strchr($image,'.'),1);//it breaks the string in part so that format can be matched
if($ext!='jpg' && $ext!='jpeg' && $ext!='png' && $ext!='gif' && $ext!='JPG' && $ext!='JPEG')
{
echo "please select image";
}
else
{
$path="gallery/".$image; //folder in which image to be saved
$action=copy($_FILES['uploadgallery']['tmp_name'],$path);//name given in input type (line72)
$query="insert into gallery (`cat_id`,`galimage`) values('$a','$image')";
$result=mysql_query($query);`enter code here`
echo "insert successfully";
}
}
?>
<option selected> -- select -- </option>';
<?php $sql = "SELECT * FROM category";
$result = mysql_query($sql);
while($row=mysql_fetch_array($result)){
echo '<option value="'.$row['cat_id'].'">'.$row['categoryname'].'</option>';
}
?>
Hope it may helps you, 希望对您有帮助,
<?php
include_once("header.php");
include ("connection.php");
if(isset($_REQUEST['ansave']))
{
$a=$_REQUEST['choosecategory'];
$image=$_FILES['uploadgallery']['name'];// name given in input type
$ext=substr(strchr($image,'.'),1);//it breaks the string in part so that format can be matched
if($ext!='jpg' && $ext!='jpeg' && $ext!='png' && $ext!='gif' && $ext!='JPG' && $ext!='JPEG')
{
echo "please select image";
}else {
$path="gallery/".$image; //folder in which image to be saved
$action=copy($_FILES['uploadgallery']['tmp_name'],$path);//name given in input type (line72)
$sqlCategory = "SELECT categoryname FROM category where cat_id='".$a."' ";
$resultCategory = mysql_query($sqlCategory);
$rowCategory=mysql_fetch_array($resultCategory);
$categoryname = $rowCategory['categoryname']; // category name
$query="insert into gallery (`cat_id`,`galimage`) values('$a','$image')";
$result=mysql_query($query);
echo "insert successfully";
}
}
?>
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