PHP - Trying To Write A Select Box In Php

hello

how would i write,

if $a has 1 or more of $b do something ?
for example

Code: [Select]

if($a >1 $b){
do something
}else{
do nothing
}


or >2
or >3

etc...

thanks

I am trying to create a query that selects results from a table called matches.  The results from this table must be based on whether or not there are results from another table called match_results.  The results from the match_results table must be based on whether or not the row has either a home_id equal to the current team id, or a visitor_id equal to the current team id.  If the current team id is equal to the home_id of the result from the match_results table, then the home_score must be greater than the visitor_score.  If the current team id is equal to the visitor_id of the result from the match_results table, then the visitor_score must be greater than the home_score. 

In short, I am trying to make a query that finds a team's match record(wins and losses).  If the result from the matches table doesn't have any results from the match_results table with the same match id, then it hasn't been played it.  I hope this makes sense.  Is there any way to do this in one single query?





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Hi all,

I have fwrite() writing to a text file called 'numbers.txt'. The text file content is "23".

I learn't fwrite() from the manual, so did:

f$fp = fopen('/opt/numbers.txt', 'w');
fwrite($fp, '1');
fclose ($fp);


I have used fwrite() to try and make the file content "123", but fwrite() writes over the content of the file, resulting in it just containing "1".

I couldn't find anywhere about writing into, rather over, and hoping someone could please help me out.

how can i find reference to  something like this "navigationID=3" or "navigationID=44" within a string and then re write it to become something like this "/en/content/3/Link_Name"?

there could be more than one reference within the string and the integer val (ie 3 and or 44) will not be known.

Any help with this would be greatly appreciated.

Hi,   sorry if there already is a topic regarding my question.. I probably didn't find the right words to describe it properly.. would appreciate a link if anybody knows   I'm making a search form for an archive and my problem is that the labeling says e.g. "1400-3500 A-C".. Now i don't want to make thousands of entrys for 1 file.. How do i find that one file while searching for a number in between, like "1420" or "B"?   Thanks in advance!   Greetings localhobo
Edited by localhobo, 13 October 2014 - 01:28 PM.

like I said I'm not very good in coding. so your help to solve this one is more than appreciate. this is what I want to do....


IF I have an image_id1 in the database please show it. other than that please show me the TBC.jpg picture.

in php/sql how this will be write.

thanks

sebastien

Hi, I have a form on my homepage which on submit sends the data to my email.  I need to know if it it possible to include in the form a read only field containing the refererid in the URL

(for instance, if someone were to have visited my homepage by entering; http://www.mypage.com/?refererid=7777, could I capture that 7777 in one of the form fields so that when the form is submitted I recieve the refererid?)

Thanks, any help appreciated...

I have a web site where users can gain points, when they use my site. For example, if the user spends $10.00 then the user will get 10 points.
In my web site there is a list of 10 retails (name). When the user clicks the link it will redirect to retailers website, where user can buy items.
How can I find out how much the user has spent on the retailer?
Thank you.

Dear All,

I wish to have 2 drop down boxes, Country Select Box and Locality Select Box. The locality select box will be affected by the value chosen in the country select box.

All is working fine except that the locality select box is not being populated.
I know that the problem is in the sql statement

WHERE country_id='$co'

because i am having an error that $co is an undefined variable.

All the rest works fine because i have replaced the $co variable directly with a number (say 98) for a particular country id and it worked fine. In what way can i define this variable $co so that it is accepted by my sql statement?

Thank you for your help in advance.

MySQL Tables indicated below:

CREATE TABLE countries(
country_id INT(3) UNSIGNED NOT NULL AUTO_INCREMENT,
country_name VARCHAR(30) NOT NULL,
PRIMARY KEY(country_id),
UNIQUE KEY(country_name),
INDEX(country_id),
INDEX(country_name))
ENGINE=MyISAM;

CREATE TABLE localities(
locality_id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
country_id INT(3) UNSIGNED NOT NULL,
locality_name VARCHAR(50),
PRIMARY KEY (locality_id),
INDEX (country_id),
INDEX (locality_name))
ENGINE=MyISAM;

Extract PHP script included below:

// connect to database
require_once(MYSQL);

if(isset($_POST['submitted']))

{

// trim the incoming data

/* this line runs every element in $_POST through the trim() function, and assigns the returned result to the new $trimmed array */
$trimmed=array_map('trim',$_POST);

// clean the data

$co=mysqli_real_escape_string($dbc,$trimmed['country']);

$lc=mysqli_real_escape_string($dbc,$trimmed['locality']);

}

?>

<form action="form.php" method="post">

<p>Country
<select name="country">
<option>Select Country</option>

<?php

$q="SELECT country_id, country_name
FROM countries";

$r=mysqli_query($dbc,$q) or
trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

while($row=mysqli_fetch_array($r))

{

$country_id=$row[0];
$country_name=$row[1];
echo '<option value="' . $country_id . '"';
if(isset($trimmed['country']) && ($trimmed['country']==$country_id))
echo 'selected="selected"';
echo '>' . $country_name . '</option>\n';

}

?>

</select>
</p>

<p>Locality
<select name="locality">
<option>Select Locality</option>

<?php

$ql="SELECT locality_id, country_id, locality_name
FROM localities
WHERE country_id='$co'
ORDER BY locality_name";

$rl=mysqli_query($dbc,$ql) or
trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

while($row=mysqli_fetch_array($rl))

{

$locality_id=$row[0];
$country_id=$row[1];
$locality_name=$row[2];
echo '<option value="' . $locality_id . '"';
if(isset($trimmed['locality']) && ($trimmed['locality']==$locality_id))
echo 'selected="selected"';
echo '>' . $locality_name . '</option>\n';

}

// close database connection
mysqli_close($dbc);

?>

</select>
</p>

<p><input type="submit" name="submit" value="Submit" /></p>

<input type="hidden" name="submitted" value="TRUE" />

</form>

Hey guys,

any suggestions how i could read and write msWord files without using the COM interface. I need to keep all the formatting intact though... any ideas?

many thanks

<?php
$a = 2;
$b = 3;
$z = fopen("test.txt", 'a+');
if($a > $b)
  {
  echo "A is bigger than B";
  }
  elseif fwrite($z, $b);
fclose($z);
?>


is there a way to get elseif to write to a file?

so if b is bigger than a it writes $b to a file?

Thats my question. Its a good practice? Gives error in a php file without html, for example?

About the library, i have a 1500lines single php file and i modualritizing it. There is any advice in general when creating a php library?

suggestions on how to write this code better
I have a form that has 2 textboxes in it. The php listed below...confirms that the passwords match, confirms that there are at least 6 characters in in, confirms that there is at least 1 number. I am doing mostly if statements with it, but I was wondering if there was a better way to write.

Code: [Select]
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>

<body>
     <form name="form1" method="post" action="<?PHP $_SERVER['PHP_SELF'] ?>">

<table width="100%" border="1">
  <tr>
    <th scope="row"> <strong>New password:  </strong>
            <input type="password" maxlength="15" name="password"></th>
<th scope="row"> <strong>New password confirmed: </strong>
            <input type="password" maxlength="15" name="confirm_password"></th>
<th scope="row"><input type="submit" name="update" value="Update Password" /></th>
  </tr>
</table>

</form>
<?PHP
//verify that the update button is pressed
if (isset($_POST['update']))
{
//verify that the passwords match
if($_POST['password'] == $_POST['confirm_password'])
{
echo "passwords match";
echo "<br>";

}
//verify that the password and password confirmation was entered
elseif ((isset($_POST['password'])) && ($_POST['confirm_password']))
{
echo "Password and confirm password set";
echo "<br>";
}
//verify that the # of characters was entered
elseif (strlen($_POST['password'])<6)
{
$num_char=strlen($_POST['password']);
echo "Please enter more characters";
echo "<br>";
}
//find out if number exsist
if (strlen($_POST['password'])>5)
{
$subject = $_POST['password'];
preg_match_all('/[0-9]/', $subject, $matches);
$count = count($matches[0]);
//echo $count;
if ($count <=1)
echo "Please enter in at least 1 number you currently have: " .$count;
}//close of verification of both boxes entered

}//close of $_POST
?>

</body>
</html>

Why doesn't it add a new line when I open up index.php

Code: [Select]
$fp = fopen('text.txt','w');
fwrite($fp, "Your IP: ".$_SERVER['REMOTE_ADDR']."\n");
fclose($fp);