[php mysql verify two rows]

Sorry, can't type....

 

$query2 = "SELECT * FROM performancecriteria WHERE project_id = $project AND year = 2010 AND ( type = 1 OR type = 2 )";

[php mysql verify two rows]

$query2 = "SELECT * FROM performancecriteria WHERE project_id = $project AND year = 2010 AND ( type = 1 OR type + 2 )";

[Very simple adding not working !]

uhhmm

This displays the correct values:

<?php
$temp1=9;
$temp2=4;
//$temp1 = $nonRBProducts[$counteri][3];
$temp1 * 1;
//$temp2 = $RBProducts[$counterr][3];
$temp2 *  1;
$temp3 = $temp1 + $temp2;

echo"temp1 = $temp1<br />temp2 = $temp2<br />temp3 = $temp3<br />";
?>

 

Are you sure the variables $nonRBProducts[$counteri][3]; and $RBProducts[$counterr][3]; contain a value?

 

And what is the purpose of attempting to multiplying by 1? If that's what you want to do because $temp2 *  1; wont do anything!

[Using the AVG fuction]

Oh yes, doh!

try changing

$query = mysql_query("SELECT  AVG(customer_service) FROM rating WHERE cust_id LIKE '$search'"); 

to

$query = mysql_query("SELECT  AVG(customer_service) FROM rating WHERE cust_id = '$search' "); 

 

[Using the AVG fuction]

what is line 27?

[Using the AVG fuction]

try

<?php// dont use short tags!!!

//connect to mysql
//change user and password to your mySQL name and password
$link_id=mysql_connect("localhost","user","password"); 
    
//select which database you want to edit
if (mysql_select_db("voogahco_microblog", $link_id));


$search="$id";

//get the mysql and store them in $result
//change whatevertable to the mysql table you're using
//change whatevercolumn to the column in the table you want to search
// Make a MySQL Connection

$query = mysql_query("SELECT  AVG(customer_service) FROM rating WHERE cust_id LIKE '$search'"); 
     
$result = mysql_query($query) or die(mysql_error());

$row=mysql_fetch_row($result);

$average=$row[0];

echo"average=$average";
?>

[filesystem - get the image filename, then insert to database.]

try

$file_name = trim($_FILES['file']['name']);// original name of file

Then

echo $file_name;

[I can't figure out what is wrong with this mail() script.]

Hi, No I don't know....

but I put the question to Google and got loads of stuff

http://www.adrogen.com/blog/email-not-sending-to-same-domain/

[I can't figure out what is wrong with this mail() script.]

If The script works on your server but not on their server, it must be a server issue. (The script works on my apache 2 / php5).  Open a support ticket!

[help witht this table-column counter please? whey doesnt it break rows properly?]

ok, try

$columns = 0;
$columnsCounter = 2;
echo "<table border='1'>
<tr>";
while ($row = mysql_fetch_array($products)) {
   echo "<td>" . $row['title']. "<br> </td>";
   $columns++;
   if ($columns == $columnsCounter) { 
       echo "</tr><tr>";  
       $columns == 0;
   }
}
   echo "</tr></table>";

[help witht this table-column counter please? whey doesnt it break rows properly?]

at first glance you have a brace too many

   echo " </td>";
   $columns++;
   if ($columns == $columnsCounter) { echo "</tr><tr>";  }
   $columns == 0;
   }

should be

   echo " </td>";
   $columns++;
   if ($columns == $columnsCounter) { echo "</tr><tr>";  
   $columns == 0;
   }

[Saturday morning PHP]

looks like you are adding it twice

  $add_member = mysql_query($insert);
  

if (!mysql_query($insert,$con))

 

you have two insert actions

 

[Not sure how to go about this....]

I'm in agreement. Best to have a table that holds the options. Easier to operate than a code for different combinations.

[Odd error with an IF statement....]

<option value="08" <?PHP IF ($fuel1 == 08) echo "selected"; ?>>Aug</option>

 

08 is not a number, it's a string

 

Change to

<option value="08" <?PHP IF ($fuel1 == "08") echo "selected"; ?>>Aug</option>

 

and it will work

 

[php sql table help]

Put each photo with text in to a div and position with  CSS

 

http://www.starvillasjavea.com/4_bed_villas.php

[Parse XML - Write to a database --- PLEASE HELP!]

you are welcome!

[sleep() not working]

You cant have html content before a header call. It will never work. Java Script will do the job for you.

[fatal error]

Read the front page of this site!!

 

http://www.phpfreaks.com/tutorial/defining-a-php-function-only-once

[Help parsing csv file]

desc is a reserved word in MYSQL you need to change the field name.

[Help parsing csv file]

Lets see the new insert statement

[Dropdown list help]

Where are the options coming from? Is the list dynamically generated? Lets see some code.

[Help parsing csv file]

NOT

$sql = "INSERT INTO $table SET category='$category', supp_pn='$supp_pn', desc='$desc', $help='help'";
mysql_query($sql)or die (mysql_error());

BUT

$sql="INSERT INTO $table (category, supp_pn, desc, help)VALUES('$category', '$supp_pn', '$desc', '$help' )";

[Select From Help]

You can't have a variable name that has a space in it!

you have the ID as a variable

Why not...

$query_Connectors= sprintf("SELECT * FROM Connectors WHERE Conn_ID = %s",
         mysql_real_escape_string($_GET['EquipID']));

[Select From Help]

this is the first error

$query_Connectors = "SELECT * FROM Connectors WHERE Conn_ID = %s", GetSQLValueString($colname_Standard Connector, "int"));

 

Space in variable name $colname_Standard Connector

 

and same error furter down.....

[Parse XML - Write to a database --- PLEASE HELP!]

I think this will solve it for you...

$reference[$name][$this->_levelsNumbering[$i][$name]] =$reference[$name][$this->_levelsNumbering[$i][$name]].$data

 

notice no space before . and $data

 

I had a similar problem.....