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PHP/My SQL search on empty fields

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PHP/My SQL search on empty fields MySQL Help forum discussing administration, SQL syntax, and other MySQL-related topics. MySQL is an open-source relational database management system (RDBMS).

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  #1  
Old October 10th, 2000, 01:01 PM
Mike Paxman Mike Paxman is offline
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Can anyone help a new PHP user?

I'm constructing a search on a MySQL database.

I have a field for thumbnail pictures, (picthumb)
but some of the entries don't have thumbnail pictures.

What I want is that when searching, if the thumbnail field is empty,
then the resulting html page pulls up an image called coming.gif
(which shows that the thumbnail is not yet available)

I'm sure this is a simple if - else line, but I can't get it right.

any help would be appreciated!
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  #2  
Old October 10th, 2000, 01:06 PM
rod k rod k is offline
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Do it in your query:

select foo,bar,if(isnull(thumb),'coming.jpg',thumb) as thumb from table....;

Assuming the field name is 'thumb' and that you are storing the actual filename.
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  #3  
Old October 10th, 2000, 01:54 PM
Mike Paxman Mike Paxman is offline
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Thanks Rod, but I don't seem to be able to get it working, have I got it in the right place?
Here's the script.
BTW the field is named (picthumb)

$result = mysql_query ("SELECT * FROM propdata1 WHERE location LIKE '$location%' AND rental LIKE '$rental%' AND purchase LIKE '$purchase%' ");

if(isnull(picthumb),'coming.gif',picthumb);
if ($row = mysql_fetch_array($result)) {

do {
printf ("<body bgcolor=#FFFFFF>");
printf ("<center><P><table border=0 cellpadding=5 cellspacing=0 width=600>n");
printf ("<tr><td colspan=2>n");
printf ("<strong>n");
printf (" %sn",$row["name"]);
printf (" - ");
printf (" %sn",$row["location"]);
printf ("</td></tr><tr><td>");
printf ("</strong><br>");
printf ("<img src=http://www.blah.com/img/");
printf ($row["picthumb"]);
printf ("</td></tr>n");
printf ("</table></P></center>n");
} while($row = mysql_fetch_array($result));
} else {print "<br><br><center><b> Sorry, no records were found! </b></center>";}


[This message has been edited by Mike Paxman (edited October 10, 2000).]
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  #4  
Old October 10th, 2000, 02:34 PM
rod k rod k is offline
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You can do it in your query or outside. The code I gave you was for inside the query. To do it the way you are trying:

<code>$result = mysql_query ("SELECT * FROM propdata1 WHERE location LIKE '$location%' AND rental LIKE '$rental%' AND purchase LIKE '$purchase%' ");


if ($row = mysql_fetch_array($result)) {
if $row[picthumb]=="" $row[picthumb]='coming.jpg';

do {
printf ("<body bgcolor=#FFFFFF>");
printf ("<center><P><table border=0 cellpadding=5 cellspacing=0 width=600>n");
printf ("<tr><td colspan=2>n");
printf ("<strong>n");
printf (" %sn",$row["name"]);
printf (" - ");
printf (" %sn",$row["location"]);
printf ("</td></tr><tr><td>");
printf ("</strong><br>");
printf ("<img src=http://www.blah.com/img/");
printf ($row["picthumb"]);
printf ("</td></tr>n");
printf ("</table></P></center>n");
} while($row = mysql_fetch_array($result));
} else {print "<br><br><center><b> Sorry, no records were found! </b></center>";}</code>
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  #5  
Old October 10th, 2000, 03:04 PM
Mike Paxman Mike Paxman is offline
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The middle line here is throwing a parse error:

if ($row = mysql_fetch_array($result));{
if $row[picthumb]=="" $row[picthumb]='coming.gif';
do {
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  #6  
Old October 10th, 2000, 05:06 PM
rod k rod k is offline
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Remove the ; from between the first if and the opening {
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  #7  
Old October 10th, 2000, 05:29 PM
Mike Paxman Mike Paxman is offline
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I've already done that, but I still get a

Parse error: parse error, expecting `'(''

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  #8  
Old October 10th, 2000, 09:14 PM
rod k rod k is offline
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Doh! Forgot the parenthesis.

if ($row[picthumb]=="") $row[picthumb]='coming.jpg';
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  #9  
Old October 11th, 2000, 04:32 AM
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You could also set a default in your table, so instead of storing NULL it will store 'coming.jpg' Then there won't be any additional logic you have to do

You can use the alter command to change it.

ALTER TABLE your_table CHANGE COLUMN your_column_name new_column_name VARCHAR(50) DEFAULT 'coming.jpg';

I'm assuming it's a varchar column, adapt to your needs. Also, have the two column names be the same if you don't want to change the name.

It's another approach...

---John Holmes...

------------------
*************************************************************
* The manual can probably answer 90% of your questions...
*
* PHP Manual. www.php.net/manual
* MySQL Manual: www.mysql.com/documentation/mysql/bychapter
*************************************************************
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  #10  
Old October 12th, 2000, 02:18 AM
Mike Paxman Mike Paxman is offline
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Thanks for the help! Problem solved.
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