input number output string how

[kamsmartx]

take this code

<select name="FileNo" id="select2" onChange="change_title();"> 
    <?  

$query="SELECT * FROM pationts"; 
$result=mysql_query($query) or die("error: " . mysql_error()); 
while($row=mysql_fetch_array($result)) 
{ 
    echo "bla"; 
?>   

<option value="<?php echo $row['FileNo'] ; ?>"><?php echo $row['FileNo'] ; ?></option>    

<? 
    }   

?> 
</select> 

he above code is for list i need to select value from it and when i select value we must change vale on the below code
Code:

<input name="PationtName" type="text" id="title"/> 

the result is :-
when i select FileNo:111 we print on PationtName: 111
this wrong we must print name of pationt who FileNo is 111

[niroshan]

when i select FileNo:111 we print on PationtName: 111

I`m assuming your using java script function for this and if so just simply use the value as patient name instead of FileNo.

<option value="<?php echo $row['patient_name'] ; ?>"><?php echo $row['FileNo'] ; ?></option>

[kamsmartx]

؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟

[niroshan]

If you can post your table structure its much easy for us to answer your question.

P.S:
If you don`t understand something please explain the issues instead of writing ????? so that would help us to explain things ...🙂

[kamsmartx]

ok show what i do
on Javascript part i write this code

function change_title(obj)
 { 
document.getElementById('title').value  = document.getElementById('select2').options[selected_index].text;}

and on body i write this code

<select name="FileNo" id="select2" onChange="change_title();">   

    <?    

$query="SELECT * FROM pationts";   

$result=mysql_query($query) or die("error: " . mysql_error());   

while($row=mysql_fetch_array($result))   

{   

    echo "bla";   

?>     

<option value="<?php echo $row['Name'] ; ?>"><?php echo $row['FileNo'] ; ?></option>      

<?   

    }     

?>   

</select> 

and

<input name="PationtName" type="text" id="title"/> 

but not print name of pationt why ??!!!!

[niroshan]

The problem is in your javascript code. Where do you pass "selected_index"??

there is a easy way of doing this,

//javascript
function change_title(ref)
{
     document.getElementById('title').value  = ref;
}


//php
<select name="FileNo" id="select2" onChange="change_title(this.value);">   

    <?    

$query="SELECT * FROM pationts";   

$result=mysql_query($query) or die("error: " . mysql_error());   

while($row=mysql_fetch_array($result))   

{   

?>     

<option value="<?php echo $row['Name'] ; ?>"><?php echo $row['FileNo'] ; ?></option>      

<?   

    }     

?>   

</select>

Hope this helps..

Regards,
Niroshan

[kamsmartx]

thanx the problem solved