我必须创建一些php页面,允许您插入和显示一些随机图像(在我的mysql DB中保存为blob)。 好吧,当我尝试它时,它给我错误:警告:mysql_fetch_array():提供的参数不是有效的MySQL结果资源。 在这里我的页面:
$n="SELECT COUNT('id_product')
FROM 'products'";
$value=mysql_query($n);
do
{
$selectionASC='SELECT id_product
FROM products
ORDER BY id_product ASC
LIMIT 1';
$selectionDESC='SELECT id_product
FROM products
ORDER BY id_product DESC
LIMIT 1';
$ASC=mysql_query($selectionASC)
or die ('Impossible execute the query <br />').mysql_error();
$DESC=mysql_query($selectionDESC)
or die ('Impossible execute the query <br />').mysql_error();
//____________________________________________________________________
$rand_n=rand(($ASC-1),($DESC+1));
//____________________________________________________________________
$selected='SELECT id_product,name, price, img
FROM products
WHERE id_product='.$rand_n;
//____________________________________________________________________
while($row=mysql_fetch_array($selected))
{
echo "Product'id:  "; echo $row[0];
echo '<br />';
echo "Name:  "; echo $row[1];
echo '<br />';
echo "Price::  "; echo $row[2];
echo '<br />';
echo "Immage: <img src='images/".$row['img']."' alt='Image'>";;
echo '<hr> <br />';
$value--;
}
}
while ($value==0)
有人可以这么好告诉我哪里错了吗?谢谢,抱歉我的英语很差!
编辑:
$allowedExts=array("gif", "jpeg", "jpg", "png");
if(isset($_POST['submit']))
{
$temp=explode(".", $_FILES["file"]["name"]);
$extension=end($temp);
if(isset($_FILES['file']['name']))
{
if(!empty($_FILES['file']['name']))
{
$directory='\www\v1.2\loaded';
$uploadfile = $directory . basename($_FILES['file']['name']);
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < 20000)
&& in_array($extension, $allowedExts))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
else
{
//$nameImage=$_FILES(["file"]["name"]);
$typeImage=$_FILES["file"]["type"];
$sizeImage=$_FILES["file"]["size"];
if (file_exists("upload/" . $_FILES["file"]["name"]))
{
echo $_FILES["file"]["name"] . " already exists. ";
}
else
{
if (move_uploaded_file($_FILES['file']['tmp_name'], $uploadfile))
{
$insert="INSERT INTO 'img'
VALUES ('','$uploadfile','$typeImage','$sizeImage')";
echo "File's extension valid, upload executed";
}
else
{
echo "Upload failed: ceck the size and the extension";
}
}
}
}
else
{
echo "Invalid file";
}
}
}
}
答案 0 :(得分:0)
不要在数组中传递img
。您必须传递索引号,因为您已使用mysql_fetch_array
所以,从代码中替换这一行,
echo "Immage: <img src='images/".$row['img']."' alt='Image'>";;
到
echo "Immage: <img src='images/".$row['3']."' alt='Image'>";
这里我猜img url是存储在第三个索引上的。请检查图像索引并替换3
的值=============================================== =================================
使用mysql_fetch_assoc
的更好方法。在这种情况下,您不需要从数据库
只需从代码中替换while循环。 它肯定会起作用
while($row=mysql_fetch_assoc($selected))
{
echo "Product'id:  "; echo $row['id_product'];
echo '<br />';
echo "Name:  "; echo $row['name'];
echo '<br />';
echo "Price::  "; echo $row['price'];
echo '<br />';
echo "Immage: <img src='images/".$row['img']."' alt='Image'>";
echo '<hr> <br />';
$value--;
}
答案 1 :(得分:0)
根据您的代码,我发现您错过了
的陈述$ selected = mysql_query($ selected);
以上
而($行= mysql_fetch_array($选择)){ ........ }
检查这应该对你有帮助。
答案 2 :(得分:0)
我认为这里的问题是$rand_n=rand(($ASC-1),($DESC+1));
,$ASC and $DESC
是资源而不是整数,相反你应该使用
$ASC = mysql_num_rows($ASC);
$DESC = mysql_num_rows($DESC);
然后你可以使用$rand_n=rand(($ASC-1),($DESC+1));
$selected='SELECT id_product,name, price, img
FROM products
WHERE id_product='.$rand_n;
$selected = mysql_query($selected);
while($row=mysql_fetch_assoc($selected))
{
echo "Product'id:  "; echo $row['id_product'];
echo '<br />';
echo "Name:  "; echo $row['name'];
echo '<br />';
echo "Price::  "; echo $row['price'];
echo '<br />';
echo "Immage: <img src='images/".$row['img']."' alt='Image'>";
echo '<hr> <br />';
$value--;
}