只有当我有两个<li>标签时才会改变图像

Question

Thanks for your timing in reading this post. Let me explain my requirement

Actually based on the IMAGENUM field value in the database, i need to display the number of images accordingly. My Updated code is as follows

while( $row = $result->fetch_assoc() ) {
            $NUM=intval($row["IMAGENUM"]);
            $imagesArr=array();
            //dynamically fill the imagesArr
            for($i=0; $i < $NUM ; $i++){
                $currKey = "IMG".($i+1);
                if(!empty($row[$currKey])){
                     array_push($imagesArr,$row[$currKey]);
                }
            } 
            $NAME= $row["NAME"];
            $QNTY=$row["QNTY"];
            $PRICE= $row["PRICE"];
            $DESC=$row["DESC"];
            $ADDDESC= $row["ADDDESC"];
            $IFLAG=$row["IFLAG"];
            $SFLAG= $row["SFLAG"];
            $CFLAG=$row["CFLAG"];
            $IMG1= $row["IMG1"];
            $IMG2=$row["IMG2"];
            $IMG3= $row["IMG3"];
            for($i = 0; $i < $NUM; $i++) {
            echo $imagesArr[$i];            
            echo '
                    <div class="single_grid">
                    <div class="grid images_3_of_2">
                        <ul id="etalage">
                            <li>
                                <a href="optionallink.html">
                                    <img class="etalage_thumb_image" src="', $imagesArr[$i], '" class="img-responsive" />
                                    <img class="etalage_source_image" src="', $imagesArr[$i], '" class="img-responsive" title="" />
                                </a>
                            </li>
                            <li>
                                <a href="optionallink.html">
                                    <img class="etalage_thumb_image" src="', $imagesArr[$i], '" class="img-responsive" />
                                    <img class="etalage_source_image" src="', $imagesArr[$i], '" class="img-responsive" title="" />
                                </a>
                            </li>
                        <ul>
                    </div>
                    </div>
            ';
            }
            }

When i echo $imagesArr[$i], i can see all the values but image is getting changed only when i have two list tags Please suggest. Thanks

Answer 1

You could push all images into an array and iterate this array afterwards.

while ($row = mysql_fetch_assoc($result)) { 
            $NUM=intval($row["IMAGENUM"]);
            $imagesArr=array();
            //dynamically fill the imagesArr
            for($i=0; $i < $NUM ; $i++){
                $currKey = "IMG".($i+1);
                if(!empty($row[$currKey])){
                     array_push($imagesArr,$row[$currKey]);
                }
            }          

            $NAME= $row["NAME"];
            $QNTY=$row["QNTY"];
            $PRICE= $row["PRICE"];
            $DESC=$row["DESC"];
            $ADDDESC= $row["ADDDESC"];
            $IFLAG=$row["IFLAG"];
            $SFLAG= $row["SFLAG"];
            $CFLAG=$row["CFLAG"];

            for($i = 0; $i < $NUM; $i++) {
              echo '
                    <div class="single_grid">
                    <div class="grid images_3_of_2">
                        <ul id="etalage">
                            <li>
                                <a href="optionallink.html">
                                    <img class="etalage_thumb_image" src="', $imagesArr[$i], '" class="img-responsive" />
                                    <img class="etalage_source_image" src="', $imagesArr[$i], '" class="img-responsive" title="" />
                                </a>
                            </li>
                        <ul>
                    </div>
                    </div>
            ';
 }

Note: Instead of using deprecated mysql_fetch_assoc(), please consider using PDO/mysqli, e.g.:

$connection = new mysqli($host,$user,$pass,$db);

$result = $connection->query("SELECT * FROM `Blogs`");


while( $row = $result->fetch_assoc() ) {
   //do your stuff here for each row
}

Furthermore, your table structure is not ideal. It would be better to store the image links separately in one table with a foreign key to the product. Then you could easily fetch all images for the product and don't need to dynamically fill the images array.