display image but shows an array

Question

I am creating a newsletter and I want to display my database image which I have inserted by name into my database. when I try to display image, but image is not displayed.

$date = $_POST['date'];

$qry_banner = "INSERT INTO `banner_tbl`(`banner_img`, `date`) VALUES ('$img_banner','$date')";
$res_banner = mysql_query($qry_banner);
$banner_tbl_id = mysql_insert_id();


$sec_id = $_POST['sectionID'];
$sec_active_image = $_POST['activeimage'];
$sec_title = $_POST['section_title'];


$total_sec = count($sec_id);

for($i=0;$i<$total_sec;$i++) 
{
$qry_section1 = "INSERT INTO `section_lt_tbl`(`banner_id`, `sectionID`, `activeimage`, `sectiontitle`) VALUES 
('$banner_tbl_id','$sec_id[$i]','$sec_active_image[$i]','$sec_title[$i]')";

 $nav_1_img = mysql_query("SELECT `activeimage` FROM `section_lt_tbl` WHERE sectionID = '1' limit 1");
 $nav_2_img = mysql_query("SELECT `activeimage` FROM `section_lt_tbl` WHERE sectionID = '2' limit 1");
 $nav_3_img = mysql_query("SELECT `activeimage` FROM `section_lt_tbl` WHERE sectionID = '3' limit 1");
 $nav_4_img = mysql_query("SELECT `activeimage` FROM `section_lt_tbl` WHERE sectionID = '4' limit 1");

$res_section1 = mysql_query($qry_section1);
$sec_tbl_id1 = mysql_insert_id();

$section_generated_id[] = $sec_tbl_id1;

}
 $nav_1_img_val = mysql_fetch_assoc ($nav_1_img);
 $nav_2_img_val = mysql_fetch_assoc ($nav_2_img);
 $nav_3_img_val = mysql_fetch_assoc ($nav_3_img);
 $nav_4_img_val = mysql_fetch_assoc ($nav_4_img);

Below is my html code where I'm trying to display image:

<td bgcolor="#dbdbdb" align="center"><table border="0" cellpadding="0" cellspacing="0" width="550" align="center">
       <tr>

       <td width="25%"><a href="#featuredstartup"><img src="uploads/<?php echo $nav_1_img_val ?>_blue.jpg" width="125px" height="134px" />
</a></td>
       <td width="25%"><a href="#healthcaredeals"><img src="uploads/<?php echo $nav_2_img_val ?>_grey.jpg" width="125px" height="134px" /></a></td>
       <td width="25%"><a href="#healthcarenews"><img src="uploads/<?php echo $nav_3_img_val ?>_grey.jpg" width="125px" height="134px" /></a></td>
       <td width="25%"><a href="#healthcaremarkets"><img src="uploads/<?php echo $nav_4_img_val ?>_grey.jpg" width="125px" height="134px" /></a></td>

       </tr>
       </table>
       </td>

you can select all images with one `mysql_query` – Abdulla Nilam Sep 22 '15 at 05:35 — Abdulla Nilam, Sep 22 '15 at 05:35

score 2 · Answer 1 · answered Sep 22 '15 at 05:34

2

Modify your img src tag to following.

<img src="uploads/<?php echo $nav_1_img_val['activeimage'] ?>_blue.jpg" width="125px" height="134px" />

mysql_fetch_assoc() will return you an associative array and you have to fetch the corresponding key value.

answered Sep 22 '15 at 05:34

Gautam

1,046
5
20

score 1 · Answer 2 · answered Sep 22 '15 at 05:33

1

You have to use $nav_1_img_val['activeimage'] instead of $nav_1_img_val because mysql_fetch_assoc is returning an array.

You should also think about moving to mysqli or PDO because mysql is deprecated.

answered Sep 22 '15 at 05:33

mario.van.zadel

2,919
14
23

yeah. definitely will move to mysqli now. do you know about any latest php-mysqli tutorial? – chintan suthar Sep 22 '15 at 05:48
@chintansuthar check http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php/14110189#14110189 not telling its complete but you may found it informative – NullPoiиteя Sep 22 '15 at 05:49
@NullPoiиteя sure i will check this link. and thank you to suggest me. – chintan suthar Sep 22 '15 at 05:51

score 1 · Answer 3 · edited Sep 22 '15 at 06:08

1

mysql_fetch_assoc returns an associative array.

You need to access the value of the array by using the name of the column as key.

 $nav_1_img_result = mysql_fetch_assoc ($nav_1_img);
 $nav_1_img_val = $nav_1_img_result['activeimage'];

edited Sep 22 '15 at 06:08

Muhammad Usman

1,366
5
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33

answered Sep 22 '15 at 05:36

Christoph Mueller

101
4

display image but shows an array

3 Answers3