Calculations in php

[2cool]

Hi

I'm tyring to do this calculation, shown below:

 

$Height='Height';

$S1Weight='S1Weight';

$result= ($Height * $Height / $S1Weight);

echo ($result);

 

Another way i tried was:

 

print ("BMI for Session 1 is: ");

print ("S1Weight FROM `main` WHERE clientID='1'" * "S2Weight FROM `main` WHERE clientID='1'" / "Height FROM `main` WHERE clientID='1'");

 

It gives me an error: 'Division by zero', There are figures in all the fields i am using ,i.e. Height and S1Weight and i am trying to individualise to to clientID. Any ideas why it throws this error?

 

Thanks in advance.

[Mchl]

$S1Weight='S1Weight';

 

You're asigning a string to a variable... how do you expect to get any results?

 

And this second method... just don't think about it anymore...

[Maq]

You also posted this in the wrong section (do not start a new one), you probably meant to put this in the "PHP Math" section.  Also, please use

tags for you code.

[2cool]

What i am trying to do is use the fields from my database (S1Weight and Height) to work out this calculation. I am a new programmer and not sure if I'm doing this the right way. I have tried many different ways but none have been successful. The code i am using now is:

 

<?php 
$query="SELECT Height FROM `main` WHERE clientD='1'"; 
$result=mysql_query($query); 
$row = mysql_fetch_assoc($result); 
echo $row['Height'];
?>
<br>
<?php
$query="SELECT S1Weight FROM `main` WHERE PatientID='1'"; 
$result=mysql_query($query); 
$row1 = mysql_fetch_assoc($result); 
echo $row1['S1Weight']; 
?>
<br>
<?php
$c=($row*$row/$row1);
echo $c;
?>

 

The first two parts display the Height and S1Weight individually but when i put it into the calculation i need i throws an error, 'Fatal error: Unsupported operand types'.

Firstly is this the correct code and if it is how could i go about resolving this error. If not, can you direct me the right way.

 

Thanks

[sasa]

<?php 
$query="SELECT Height FROM `main` WHERE clientD='1'"; 
$result=mysql_query($query); 
$row = mysql_fetch_assoc($result); 
echo $h = $row['Height'];
?>
<br>
<?php
$query="SELECT S1Weight FROM `main` WHERE PatientID='1'"; 
$result=mysql_query($query); 
$row1 = mysql_fetch_assoc($result); 
echo $w = $row1['S1Weight']; 
?>
<br>
<?php
$c=$w!=0 ? ($h*$h/$w) : 'S1Weight is zero!';
echo $c;
?>

 

[2cool]

Thanks sasa, I have entered the amended code but it gives this error:

 

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource

 

[lonewolf217]

that probably means your query is failing

 

<?php
$result=mysql_query($query) or die (mysql_error());
?>

[Maq]

Just a wild guess, but did you mean "clientID" rather than "clientD"?  (notice the I)

[2cool]

Silly me,  I think I've spent too much time on it...I have corrected the spelling and it works fine now. Thank you all for the help.