如何从其他表PHP - MySQL中获取所有图像名称

Question

配对使用其他表数据我遇到了一些问题。请看我的代码：

$sql = mysqli_query($con, "SELECT * FROM properties WHERE category like '$property_for' AND property_type like '$property_type' 
            AND search_radius >= '$area' AND price >= '$price_min' 
            AND price <= '$price_max'  AND bed_rooms like '$beds' AND bath_rooms like '$baths' 
            ORDER BY date_added DESC") or die(mysqli_error($con));
            $productCount = mysqli_num_rows($sql); 

            if ($productCount > 0) 
            {

                while($row = mysqli_fetch_array($sql))
                { 
                         $pid = $row["property_id"];
                         $agent_id = $row["agent_id"];
                         $property_name = $row["property_name"];
                         $category = $row["category"];
                         $location = $row["location"];
                         $property_type=$row["property_type"];
                         $price = $row["price"];
                         $bed_rooms = $row["bed_rooms"];
                         $bath_rooms = $row["bath_rooms"];
                         $commercial_type = $row["commercial_type"];
                         $area = $row["area"];
                         $address = $row["address"];
                         $description = $row["description"];
                         $date_added = $row["date_added"];
                         $date_added = strftime("%b %d, %Y", strtotime($row["date_added"]));

                            $imagesql = mysqli_query("SELECT * from tbl_images WHERE pid = '$pid'");
                            while($row = mysqli_fetch_array($imagesql))
                            {
                            $image = row['image_name'];
// Here there are totally 3 image_names. I want to fetch all the image names
                            }                           


                }

            } 

我在这里使用两个名为properties和tbl_images的表。在properties中，我插入了类别，价格，大小等所有数据。都是单一的价值观。但在tbl_images中，我插入了3 image_names，引用了properties表的product_id。所以，现在我需要获取properties表中的所有数据以及image_names表中的所有3 tbl_images，并在property_id表中引用properties。这是我的问题。所以，Mates帮助我......我认为Array会帮助这个脚本。但我不知道在哪里以及如何做到这一点。所以请帮助我。

Answer 1

$imagesql = mysqli_query("SELECT * from tbl_images WHERE pid = '$pid'");

$images = array();

while($i = mysqli_fetch_array($imagesql)) {

    $images[] = $i['image_name'];

}

var_dump($images); //will contain an array of all images

Answer 2

There is two while loops using the variable with the same name $row in your code. The variable in the second loop will override the variable in the first one. Change $row in the second loop to $row2.

Next, you can use two-dimensional array of images, i.e. $images[$pid][] instead of $image. This way is preferred if the number of images per property in tbl_images may increase in future.

while($row2 = mysqli_fetch_array($imagesql))
{
    $image[$pid][] = row['image_name'];
}     

Or, if the number of images in tbl_images is fixed and equals to 3 per property and will not increase in future, you can merge two queries into the big one. To do it you should add new column to the tbl_images, for example type with values 1, 2 and 3 for each image. Here is code:

Table tbl_images structure:

id  image_name   created_time    pid    type
1   first        2015-01-01      1      1
2   second       2015-01-02      1      2
3   third        2015-01-03      1      3
4   alpha        2015-01-04      1      1
5   beta         2015-01-05      1      2
6   gamma        2015-01-06      1      3
...

Query

$sql = mysqli_query($con, "SELECT a.*, b.image_name as image1, c.image_name as image2, d.image_name as image3 FROM properties WHERE a.category like '$property_for' AND a.property_type like '$property_type' AND a.search_radius >= '$area' AND a.price >= '$price_min' AND a.price <= '$price_max' AND a.bed_rooms like '$beds' AND a.bath_rooms like '$baths' LEFT JOIN `tbl_images` b ON b.pid = a.property_id AND b.type = 1 LEFT JOIN `tbl_images` c ON c.pid = a.property_id AND c.type = 2 LEFT JOIN `tbl_images` d ON d.pid = a.property_id AND d.type =  ORDER BY a.date_added DESC") or die(mysqli_error($con));

if ($productCount > 0) 
{
    while($row = mysqli_fetch_array($sql))
    { 
        $pid = $row["property_id"];
        $agent_id = $row["agent_id"];
        $property_name = $row["property_name"];
        $category = $row["category"];
        $location = $row["location"];
        $property_type=$row["property_type"];
        $price = $row["price"];
        $bed_rooms = $row["bed_rooms"];
        $bath_rooms = $row["bath_rooms"];
        $commercial_type = $row["commercial_type"];
        $area = $row["area"];
        $address = $row["address"];
        $description = $row["description"];
        $date_added = $row["date_added"];
        $date_added = strftime("%b %d, %Y", strtotime($row["date_added"]));

        $image1 = $row['image1'];
        $image2 = $row['image2'];
        $image3 = $row['image3'];
    }
}