phpretard

February 15, 2009

Believe me I know about the headers.

I think I'll just try to deal with the permissions issue and upload it to a folder.

Thanks

February 15, 2009

I can't modify the headers without error...

February 15, 2009

I can't really modify the headers...

I would like to display it as an <img src="DBInfo"> tag instead.

Is this even possible?

The code below displays it fine but I am trying to display it as the website header and it give header errors

<?

...connect

header("Content-length: $size");
header("Content-type: $type");
header("Content-Disposition: attachment; filename=$name");
echo $content;

?>

I would upload it straight to a file but... I have permission to Temp folder problems

February 13, 2009

Thank the human works great!

Although I am actually from Gallgamar.

February 13, 2009

If there are no errors on the form submitted I get PHP errors.

How can I change this to first check if the array is set before running this code?

foreach ($errormessage as $error){
echo $error;
}

Thanks!

February 12, 2009

Thank you for your help!

February 12, 2009

Someone on http://us3.php.net/setcookie said:

I found out recently that assigning FALSE to a cookie will destroy it.

I thought it might interest some of you.

look at that link anyways for some ideas

How I assign "False" to a cookie named "pi"?

February 12, 2009

I set this cookie and need to later destoy it.

<?

I SET IT WITH THIS...

$expire=time()+60*60*24*30;

setcookie("pi", "PI-1234456832", "$expire"); // Works great

I AM TRYING TO KILL IT WITH THIS...

setcookie("pi", "", time()-3600); // Doesn't work

?>

Please help me cookie monster!

February 12, 2009

Thank you!

February 12, 2009

I have a text field that is allowed numbers and text but it should contain at least 7 numbers.

This is how I would validate for 7 number only but I need at least 7 numbers and text optional.


<?
if (($var=='')||(strlen($var) < 7)||(!is_numeric($var))){

$message="Missing Numbers";

}
?>

Thank you!

February 9, 2009

<?php

$con = mysql_connect("localhost","root","b123dw9WHT3");

if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
  
mysql_select_db("company", $con);

$result = mysql_query("SELECT coName FROM contact WHERE coID='1' ");

while($row = mysql_fetch_array($result))
  {
  $info=$row['coName'];
  
  echo "<input tpe="text" value="$info" />
  }
  
mysql_close($con);
?>

February 9, 2009

how can turn this into dynamic code without typing all the options?

I would like:

$start_year = date('Y');

$end_year = $start_year -80 years

<select name="DateOfBirth_Year">
<option> - Year - </option>
<option value="2004">2004</option>
<option value="2003">2003</option>
<option value="2002">2002</option>
<option value="2001">2001</option>
<option value="2000">2000</option>
<option value="1999">1999</option>
<option value="1998">1998</option>
<option value="1997">1997</option>
<option value="1996">1996</option>
<option value="1995">1995</option>
<option value="1994">1994</option>
<option value="1993">1993</option>
<option value="1992">1992</option>
<option value="1991">1991</option>
<option value="1990">1990</option>
<option value="1989">1989</option>
<option value="1988">1988</option>
<option value="1987">1987</option>
<option value="1986">1986</option>
</select>

February 8, 2009

I am trying to display links on a page read from my DB.

I would like them to display:

CAT 1

Link 1

Link 2

Link 3

etc...

CAT 2

Link 1

Link 2

Link 3

etc...

I know the code below is silly but it's to illustrate my needs.

The DB rows are setup as is in the code...the echo is silly

<?
$result_links = mysql_query("SELECT * FROM links ORDER by cat");
while($row = mysql_fetch_array($result_links))
 {
 $id=$row['id'];
 $cat=$row['cat'];
 $url=$row['url'];
 $display=$row['display'];

 echo"<h4>Category 1</h4>";
 echo"all the links assc. with category 1";

 echo"<h4>Category 2</h4>";
 echo"all the links assc. with category 2";

 echo"<h4>Category 3</h4>";
 echo"all the links assc. with category 3";

 echo"<h4>Category 4</h4>";
 echo"all the links assc. with category 4";
 }

?>

Thanks PHP FREAKS!

January 28, 2009

		$find = strtoupper($find); 
	$find = strip_tags($find); 
	$find = trim ($find); 
	$find = explode(" ", $find);

	//Now we search for our search term, in the field the user specified 
	foreach ($find as $term){
	$data = mysql_query("SELECT * FROM cases WHERE (casetype LIKE '%$term%') OR (caseinfo LIKE '%$term%') ");
	}
	$matchescount=mysql_num_rows($data);
	$find = implode(" ", $find);

January 28, 2009

I have put this question on 3 forums with no answer.

Is the question confusing or the code?

January 28, 2009

SCENARIO 1:

"I need to Find Two words" IS IN THE DB TABLES --|casetype|-- AND --|caseinfo|--.

If I search for "Find" result returned

If I search for "Two" result returned

Problem:

If I search for "Find Two" no result.

SCENARIO 2:

"I need to Find" IS IN THE DB TABLE --|casetype|--

"Two words" IS IN THE DB TABLE --|caseinfo|--

If I search for "Find" result returned

If I search for "Two" result returned

Problem:

If I search for "Find Two" no result.

January 28, 2009

I have a site search but for somes reason it is limited to a one word search.

Search1: If I search for "Find" or "Two" it works great.

Search2: If I were to search for "Find Two" It returns nothing yet I know both are in the DB (based on Search1).

$find="ONEWORD" //the search return results
$find="TWO WORDS" //the search return nothing and I know the info is there.

$find = strtoupper($find); 
$find = strip_tags($find); 
$find = trim ($find);

$data = mysql_query("SELECT * FROM cases WHERE (casetype LIKE '%$find%') OR (caseinfo LIKE '%$find%') ");

while($result = mysql_fetch_array($data)) 
{ 
$id=$result['id'];
echo "<h1>".$result['casetype']."</h1>"; 
echo "<hr />";
$caseinfo=$result['caseinfo'];
$caseinfofull=$result['caseinfo'];
echo "$caseinfofull";
echo  $shortdesc = substr_replace($caseinfo, '', 400, -1) . "..."; 
echo "<br>"; 
}

January 28, 2009

help?

January 27, 2009

help?

January 27, 2009

Basically the same as my first question but I noticed if I search with more than one word it returns nothing.

$find="ONE"; // WORKS GREAT
$find="TWO WORDS"; // NO RESULTS ... I KNOW THERE IN THE DB THOUGH

$find = strtoupper($find); 
$find = strip_tags($find); 
$find = trim ($find); 

$data = mysql_query("SELECT * FROM cases WHERE (casetype LIKE '%$find%') OR (caseinfo LIKE '%$find%') ");
while($result = mysql_fetch_array($data)) 
{ 
echo "<h1>".$result['casetype']."</h1>"; 
echo "<hr />";
$caseinfo=$result['caseinfo'];
$caseinfofull=$result['caseinfo'];
echo $caseinfofull;
echo $shortdesc = substr_replace($caseinfo, '', 400, -1) . "; 
echo "<br>"; 
echo "<br>"; 
}

January 27, 2009

Bouya!

January 27, 2009

This only displays one result and I can't figure out why...

There are 2 fields in the DB I am trying to search through and look for the keyword typed in the search field:

FIELD 1:casetype VARCHAR

FIELD 2:caseinfo TEXT

I can get it search and display all when search one field at a time but it only dispalys one result if I try to search both fields.

$find=$_POST['find'];

$data = mysql_query("SELECT * FROM cases WHERE (casetype or caseinfo LIKE'%$find%') ");
	while($result = mysql_fetch_array($data)) 
	{ 
	$id=$result['id'];
	echo "<h1>".$result['casetype']."</h1>"; 
	echo "<hr />";
	$caseinfo=$result['caseinfo'];
	$caseinfofull=$result['caseinfo'];
	echo"<div id='$id' style='display:none'>$caseinfofull [<a href='javascript:void(0)' onclick=\"Hide('$id'), Show('partial$id')\">close</a>]</div>";
	echo "<div id='partial$id' style='display:block'>".$shortdesc = substr_replace($caseinfo, '', 400, -1) . "... <a href='javascript:void(0)' onclick=\"Show('$id'), Hide('partial$id')\">read more</a></div>"; 
	echo "<br>"; 
	echo "<br>"; 
	}

January 26, 2009

Thank you...!

January 26, 2009

Thank you for the response.

I understand wht you did there.

what I need is the results of what you did to go through the databse and send an email for each query match.


while($row = mysql_fetch_array($result))
{
  $email=$row['email'];
  $billto=$row['bill_to'];
  $phone=$row['phone'];

  echo "$email<br />";
  echo "$billto<br />";
  echo "$phone<br />";
  echo "<hr />";
}

   foreach (($email  $billto  $phone) as $mailto){

   email script

   }

}

Am I making any sense?

January 26, 2009

Each row in my DB contains -- email | phone | bill_to ---

I would like to send an email to each person with the information specific to them.

The code below does display the information but I don't believe it associates them.

So each email would read:

-------------------------------------------

Dear Person I am emailing,

Here is the information we have on file:

Email Address: email[at]email.com

Bill-To Number: 123456

Phone Number: 123-456-7890

I hope you got this,

Anthony

ps. if you did get this thank you phpfreaks.

-------------------------------------------

 


while($row = mysql_fetch_array($result))
  {
  $email[]=$row['email'];
  $billto[]=$row['bill_to'];
  $phone[]=$row['phone'];
  }
  
  foreach($email as $mailto){
  echo "$mailto<br />"; 
  }
  foreach($billto as $billnumber){
  echo "$billnumber<br />"; 
  }
  foreach($phone as $telephone){
  echo "$telephone<br />"; 
  }

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phpretard

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Image From Blob?

Image From Blob?

Image From Blob?

[SOLVED] Check if Array Exists

[SOLVED] Check if Array Exists

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[SOLVED] Validate Minimum Of Numbers In A Test Filed

[SOLVED] Validate Minimum Of Numbers In A Test Filed

PHP & MySQL HELP!

[SOLVED] Birthyear + / -

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[SOLVED] Search DB Help

[SOLVED] Search DB Help

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[SOLVED] Search DB Help

[SOLVED] Search DB Help

[SOLVED] combine foreach values

[SOLVED] combine foreach values

[SOLVED] combine foreach values

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