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Display multiple images from directory with while loop

Posted on 2008-11-02
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Last Modified: 2012-06-21
Can't get this simple loop to display all images in directory - only first one.
<?php

include_once( $_SERVER['DOCUMENT_ROOT'].'/includes/php/databaseConfig.php' ); 

$query = "SELECT id, img_path FROM test1";

$result = mysql_query($query) or die(mysql_error());

if (mysql_num_rows($result) > 0) {

while($row = mysql_fetch_array($result)) 

{

	echo '<p><img src="upimg/'.$row['img_path'].'></p>';

 

    }

}

?>

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Question by:shareea
  • 7
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12 Comments
 
LVL 27

Expert Comment

by:yodercm
Comment Utility
The images in the directory is not what you're displaying, you're displaying based on what's in your table test1.

How many rows in that table?
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Author Comment

by:shareea
Comment Utility
Have been trying several files out - eyeballs getting tired. Yes, seems there's only 1 row in that table but have several images in the directory. Here's my input form:
<?php

include_once( $_SERVER['DOCUMENT_ROOT'].'/includes/php/databaseConfig.php' );      
 

$max_no_img=3; // Maximum number of images value to be set here
 

echo "<form method=post action=addimgck.php enctype='multipart/form-data'>";

echo "<table border='0' width='400' cellspacing='0' cellpadding='0' align=center>";

for($i=1; $i<=$max_no_img; $i++){

echo "<tr><td>Images $i</td><td>

<input type=file name='images[]'></td></tr>";

}
 

echo "<tr><td colspan=2 align=center><input type=submit value='Add Image'></td></tr>";

echo "</form> </table>";
 

while(list($key,$value) = @each($_FILES[images][name]))

{

if(!empty($value)){   // this will check if any blank field is entered

$filename = $value;    // filename stores the value

$add = "upimg/$filename";   // upload directory path is set

//echo $_FILES[images][type][$key];     // uncomment this line if you want to display the file type

// echo "<br>";                             // Display a line break

copy($_FILES[images][tmp_name][$key], $add);     //  upload the file to the server

chmod("$add",0777);                 // set permission to the file.

$query = "INSERT INTO test1 (img_path) VALUES ('$filename')";

mysql_query($query);

	}

}

?>

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LVL 27

Expert Comment

by:yodercm
Comment Utility
The loop you originally posted is displaying based on what rows are in the table, so if there is only one row, then you will only see one image, no matter what is in the directory.

As for the rest -- how you get the file path/names into the table -- I'm not knowledgable about that part.  Hopefully someone else who knows file handling in php will pop in and help.
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Author Comment

by:shareea
Comment Utility
Yeah - seem to be missing the part where filedata is converted to jpeg. Thanks for your help all the same.
0
 
LVL 27

Expert Comment

by:yodercm
Comment Utility
The best technique is to put some echo statements into your code at strategic places and see exactly where the problem is occurring.  This will help you sort it out.
0
 

Author Comment

by:shareea
Comment Utility
This snippet worked to display all images located in a directory named upimg:
<?php
 

$dirname = "upimg/";

$images = scandir($dirname);

$ignore = Array(".", "..");

foreach($images as $curimg){

if(!in_array($curimg, $ignore)) {

echo "<img src='upimg/$curimg' /><br>\n";

};

}     

?>

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LVL 27

Accepted Solution

by:
yodercm earned 500 total points
Comment Utility
OK, so replace the echo with the database INSERT, and then see what you get into the table.  Be sure to follow the query with

echo mysql_error();

so you know if an INSERT failed.
0
 
LVL 27

Expert Comment

by:yodercm
Comment Utility
Then just keep doing this, step by step, until you find where the process isn't doing what you expect.  I'll check in again tomorrow morning :)
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Author Comment

by:shareea
Comment Utility
Files are being uploaded to the directory and displaying but only first image is being entered into database table. Need to loop the query. Any suggestions?
<?php

include_once( $_SERVER['DOCUMENT_ROOT'].'/includes/php/databaseConfig.php' );      
 

$max_no_img=3; // Maximum number of images value to be set here

?>

<form method=post action="<?php echo $_SERVER['PHP_SELF']; ?>" enctype='multipart/form-data'>

<table border='0' width='400' cellspacing='0' cellpadding='0' align=center>

<?php for($i=1; $i<=$max_no_img; $i++){

echo "<tr><td>Images $i</td><td>

<input type=file name='images[]' ></td></tr>";

}
 

echo "<tr><td colspan=2 align=center><input type=submit value='Add Image'></td></tr>";

echo "</form> </table>";
 

while(list($key,$value) = @each($_FILES[images][name]))

{

if(!empty($value)){   // this will check if any blank field is entered

$filename = $value;    // filename stores the value

$add = "upimg/$filename";   // upload directory path is set

//echo $_FILES[images][type][$key];     // uncomment this line if you want to display the file type

// echo "<br>";                             // Display a line break

copy($_FILES[images][tmp_name][$key], $add);     //  upload the file to the server

chmod("$add",0777);                 // set permission to the file.
 

$query = "INSERT INTO test1 (img_path) VALUES ('$filename')";

mysql_query($query);

}

}

//view files in directory

$dirname = "upimg/";

$images = scandir($dirname);

$ignore = Array(".", "..");

foreach($images as $curimg){

if(!in_array($curimg, $ignore)) {

echo "<img src='upimg/$curimg' width='200px'/><br>\n";

};

}     

?>

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Author Comment

by:shareea
Comment Utility
And after adding echo mysql_error(); get following error:

Duplicate entry '0' for key 1
0
 

Author Comment

by:shareea
Comment Utility
OK - worked it out.

$query = "INSERT INTO test1 (id, img_path)  VALUES ('$key','$filename')";
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Author Closing Comment

by:shareea
Comment Utility
Thanks! Mysql error() helped just like you said.
0

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