1

hello guys im new here so please be kind with me, please be patience cause Im new in PHP.
now here's my problem I really miss-up my code. I want to create a form with 4 dropdownlist (diagnosis, cause, medicine, and treatment) this drop-downs are dynamic taking there items from my database... cause, medicine, and treatment items are dependent on diagnosis value..

and in diagnosis there is a static item which is other with a value of other. now I want a textbox to appear every time other is selected in my diagnosis dropdown....

//by: lifeworks

<SCRIPT language=JavaScript>
window.onload = initForm;
 
function initForm(){
	document.getElementById('selectedDiag').selectedIndex = 0;
	document.getElementById('selectedDiag').onchange = showHideBox;
}
 
function showHideBox(){
 
	var whichForm = document.getElementById('selectedDiag');
	var theForm = whichForm.options[whichForm.selectedIndex].value;
	if(whichForm.options[whichForm.selectedIndex].value == 'other'){
		document.getElementById('selectedOther').style.display='';
	}
	else{
		document.getElementById('selectedOther').style.display='none';
	}
 
}

</script>

this is for the textbox to appear...

<SCRIPT language=JavaScript>
function reload(form)
{
var val=form.selectedDiag.options[form.selectedDiag.options.selectedIndex].value;
self.location='updateMedicalRecords.php?selectedDiag=' + val ;
}  	

</script>

this code does not functional since I put function showHideBox()

echo "<form id=f1 method=post name=f1 action=''>";

echo "<select id='selectedDiag' name='selectedDiag' onchange=\"reload(this.form) && showHideBox()\"><option value=''>Select diagnosis</option>"."<option value='other'>other</option>";
while($noticia2 = mysql_fetch_array($quer2)) { 
if($noticia2['diagnosisId']==@$diag){echo "<option selected value='$noticia2[diagnosisId]'>$noticia2[diagnosisName]</option>"." "."<BR>";}

 else {
	echo "<option value=$noticia2[diagnosisId]>$noticia2[diagnosisName]</option>";
}
}

echo "</select>";

//////////        Starting of second drop downlist /////////

echo "<select name='cause'>
<option value=''>Select cause</option>";
while($noticia = mysql_fetch_array($quer)) { 
echo  "<option value='$noticia[causeId]'>$noticia[cause]</option>";
}
echo "</select>";

//////////        Starting of third drop downlist /////////
echo "<select name='medicine'>
<option value=''>Select medicine</option>";
while($noticiamed = mysql_fetch_array($mediquery)) { 
echo  "<option value='$noticiamed[medicineId]'>$noticiamed[medicineName]</option>";
}
echo "</select>";

echo "<input type=submit value=Submit>";

echo "</form>";

and also the third dropdown has no item appearing, I dont know if its in my PHP code of in query..

<input name='txt' type='text' id='selectedOther' style='display:none' />

please help me guys.. and more power

2
Contributors
4
Replies
6
Views
5 Years
Discussion Span
Last Post by kalaban
0

Hi!

I've read that you are just starting to learn PHP! Good :)

It makes it easier to read your code if you have a few coding guidelines:

>> Indent and comment your code, keeping statements and function calles seperated and clarified:

/* Doing something: */
doSomething();

/* Doing something else: */
doSomethingElse();

/* Indicates whether something needs to be shown: */
$shown = true;

/* Only if it needs to be shown: */
if ($shown == true) {
     
     /* Showing something: */
     echo "Something";
     
}

/* Doing a loop: */
while ($a == 0) {

     //.... Code block ....

}

// I hope you get the picture by now,
// because typing this isn't really helping you at all with your problem...

>> Also, you might want to read up on how to properly write HTML (watch the use of quotes and ';'):

<img src='myimage.png' border='0' onclick='doSomething("StringArgument");' />

And as a final check, you can always go to the guys that design HTML: http://validator.w3.org/

>> Oh and one last thing, don't use old syntax for your HTML:

<script type='text/javascript'>
function reload(form)
{
  var val = form.selectedDiag.options[form.selectedDiag.options.selectedIndex].value;
  self.location='updateMedicalRecords.php?selectedDiag=' + val ;
}  
</script>

I think you will find your answer when you've cleaned up your code.

~G

0

sir Graphix thank you for your response and tips..
I dont know if this is right but I just replace the quotes with ' when Im coding HTML inside <?php tag. I really mess-up this one big time Im working for this almost one month and I cant find the answer. Is there a way that I can write those functions combined? I'm into cleaning up my code but those javascript giving me a hard time. xp

0

In my opinion, the three javascript function initForm and showHideBox are unnessecary. When the page reloads, the option that is chosen is already selected (so initForm is obsolete) and using a simple if condition, you can render showHideBox obsolete too!

I'il explain with some code:

//
// As you have not given all your code I assume a few things:
//    > $diag is the ID of the selected diagnostic ($_GET['selectedDiag'])
//    > The query $quer2 retrieves all the diagnostic dropdown values
//    > The query $quer retrieves all causes WHERE diagnosticId = $diag (and same with $mediquery)


/* Opening form: */
echo "<form method='post' name='f1' action=''>";

/* Opening diagnostic dropdown with default option: */
echo "
<select id='selectedDiag' name='selectedDiag' onchange='reload(this.form)'>
<option value=''>Select diagnosis</option>
";

while($noticia2 = mysql_fetch_array($quer2)) { 
	if($noticia2['diagnosisId'] == $diag) {
	  echo "<option selected='selected' value='".$noticia2['diagnosisId']."'>".$noticia2['diagnosisName']."</option>";
	} else {
	  echo "<option value='".$noticia2['diagnosisId']."'>".$noticia2['diagnosisName']."</option>";
	}
}

/* Closing diagnostic dropdown with 'other' option: */
echo "<option value='other'>Other</option>
</select>";

/* Only if the diagnostic is set: */
if (isset($diag)) {

/******************************************************************************
 * I'il let you correct this part yourself: 
*/
	echo "<select name='cause'>
	<option value=''>Select cause</option>";
	while($noticia = mysql_fetch_array($quer)) { 
	echo  "<option value='$noticia[causeId]'>$noticia[cause]</option>";
	}
	echo "</select>";

	//////////        Starting of third drop downlist /////////
	echo "<select name='medicine'>
	<option value=''>Select medicine</option>";
	while($noticiamed = mysql_fetch_array($mediquery)) { 
	echo  "<option value='$noticiamed[medicineId]'>$noticiamed[medicineName]</option>";
	}
	echo "</select>";

/*********************************************************************************/

}

/* Showing submit button: */
echo "<input type='submit' name='submit_button' value='Submit'>";

/* Closing form: */
echo "</form>";

I think this will help you make the four dropdown work.

~G

Edited by Graphix: n/a

0

Sir Graphix;
thank you for this wonderful post.
and all your guess bout my code is correct, how did you do that?
now the four dynamic dropdownlist is solve,

here's the code I come up with

function reload(form) {
		var val=form.selectedDiag.options[form.selectedDiag.options.selectedIndex].value;
		if (val == 'other'){
		document.getElementById('otherDiag').style.display='';
		document.getElementById('otherCause').style.display='';
		document.getElementById('otherMedicine').style.display='';
		document.getElementById('otherTreat').style.display='';

		} else {
			<!--var val=form.selectedDiag.options[form.selectedDiag.options.selectedIndex].value;-->
			self.location='updateMedicalRecords.php?studentsId=<?php echo $getstudentsId; ?>&selectedDiag=' + val ;
			}
}

I just play with it, and I need to modify the script for I need to put other option in every drop down and a corresponding textbox..

and bye the way sir what is the proper way of putting the selected value from drop down list into php variable, and insert it to database?

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