Dear friends,
I tried to modify submit if elseif statement. That didn't work.I have pasted
code of form and output with error.
Any comments on erros, please.
Asif
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Notice: Undefined variable: submit in C:\webroot\display.php on line 16
Notice: Undefined variable: PHP_SELF in C:\webroot\display.php on line 19
1
Treatment of Hypertension
a)loop diuritics
b)clonazepam
c)benzodizap
d)lorazepam
2
Thrombolytic treatment of Myocardial infarction
a)streptokinase
b)diltiazm
c)aspirin
d)heparin
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//code for above output and error
<HTML><body bgcolor=""><?php
// open the connection
$conn = mysql_connect("localhost", "", "");
// pick the database to use
mysql_select_db("testDB",$conn);
// create the SQL statement
$sql = "SELECT * FROM testTable";
// execute the SQL statement
$result = mysql_query($sql, $conn) or die(mysql_error());
//go through each row in the result set and display data
if (!$submit) {
echo "<form method=post action=$PHP_SELF>";
echo "<table border=0>";
while ($newArray = mysql_fetch_array($result)) {
// give a name to the fields
$id = $newArray['id'];
$testField = $newArray['testField'];
$testFielda = $newArray['testFielda'];
$testFieldb = $newArray['testFieldb'];
$testFieldc = $newArray['testFieldc'];
$testFieldd = $newArray['testFieldd'];
//echo the results onscreen
echo
"<tr><td colspan=4><br><b>$id</b></td></tr>";
echo
"<TD>$testField<input type=hidden name=$testField value=\"$testField\"> <br>
</td></tr>
<TD>a)<input type=radio name=$testFielda
value=\"$testFielda\">$testFielda</td></tr>
<TD>b)<input type=radio name=$testFieldb
value=\"$testFieldb\">$testFieldb</td></tr>
<TD>c)<input type=radio name=$testFieldc
value=\"$testFieldc\">$testFieldc</td></tr>
<TD>d)<input type=radio name=$testFieldd
value=\"$testFieldd\">$testFieldd</td></tr>";
}
echo "</table>";
echo "<input type='submit' value='See how you did' name='submit'>";
echo "</form>";
}
elseif ($submit)
?>
</html>
