if (isset($_POST["submit"])){
$oride='';
$count = "25";
$origin = $_POST["origin"];
$destinataion = $_POST["destination"];
$oride = ($destination = $_POST["destination"] - $origin= $_POST["origin"]);
switch (true) {
case ($count<="0"):
echo "invalid";
break;
case ($count==="15"):
echo $count;
break;
case ($count==="16"):
$total = $count + "1";
echo $total;
break;
default:
echo "hello";
} }
The code will compute 1st then execute switch depending on what is the result of the computation. I tried if else but it will be too long because the case will go up to 130.
You must use the var $count in switch statement and the constant in case this way
switch ($count) {
case "0" :
echo "invalid";
break;
case "15":
echo $count;
break;
case "16":
$total = $count + "1";
echo $total;
break;
default:
echo "hello";
break;
}
You have to provide an expression to the switch statement, while the case statements are just "versions" of the result of that expression. The only thing you can NOT do directly is the "<= 0" expression, but you can work around it:
if (isset($_POST["submit"])){
$oride='';
$count = "25";
$origin = $_POST["origin"];
$destinataion = $_POST["destination"];
$oride = ($destination = $_POST["destination"] - $origin= $_POST["origin"]);
// --- normalize $count:
$count = $count <= 0 ? 0 : $count;
// use $count as expression:
switch ($count) {
case 0:
echo "invalid";
break;
case "15":
echo $count;
break;
case "16":
$total = $count + "1";
echo $total;
break;
default:
echo "hello";
} }
Related
Hy,
I got switch case inside function when but when i call it, i got error Undefined Variable and i don't know why (i use PHP 8)
private function getIDFromBrand() {
switch ($this->brand) {
case "Niky":
$id = 1;
break;
case "Pumo":
$id = 4;
break;
case "Coke":
if ($this->typecoke== 0) {
$id = 2;
} else {
if ($this->typecoke== 1) {
$id = 3;
}
}
break;
case "Tomato":
$id = 5;
break;
case "Riles":
$id = 6;
break;
case "TEST":
$id = 7;
break;
}
return $id; // Error Undefined variable $id
}
When i declare $id at the top of my function, like
$id = null
or
$id = 0
The switch doesn't update it, so it will return null or 0, it will return the declared value.
Your switch statement has no default branch, so if $this->brand is, say, "Stack Overflow", it will not run any of the statements, and $id will never be set.
See the PHP manual for the switch statement:
A special case is the default case. This case matches anything that wasn't matched by the other cases.
Similarly, if $this->brand is "Coke", but $this->typecoke is, say, 42, it will not match either of the conditions in that branch.
switch ($this->brand) {
case "Niky":
$id = 1;
break;
case "Pumo":
$id = 4;
break;
case "Coke":
if ($this->typecoke== 0) {
$id = 2;
} elseif ($this->typecoke== 1) {
$id = 3;
} else {
$id = -1; // WAS PREVIOUSLY NOT SET
}
break;
case "Tomato":
$id = 5;
break;
case "Riles":
$id = 6;
break;
case "TEST":
$id = 7;
break;
default:
$id = -1; // WAS PREVIOUSLY NOT SET
break;
}
Going immediately to the point:
in this code I expect the default case:
<?php
$a = 0;
switch ($a) {
case "one":
echo "one";
break;
case "two":
echo "two";
break;
default:
echo "default";
break;
}
?>
I get one instead.
here expect the zero case ("two"):
<?php
$a = 0;
switch ($a) {
case "one":
echo "one";
break;
case 0:
echo "two";
break;
default:
echo "default";
break;
}
?>
I get one instead.
That appens only with zero value because here I get correctly default:
<?php
$a = 1;
switch ($a) {
case "one":
echo "one";
break;
case "two":
echo "two";
break;
default:
echo "default";
break;
}
?>
but here I get correctly zero:
<?php
$a = 0;
switch ($a) {
case 1:
echo "one";
break;
case 0:
echo "zero";
break;
default:
echo "default";
break;
}
?>
Why?
I have this PHP function for changing color between 2 numbers:
function color_switch($number){
switch (true){
case $number == range(1 , 3):
$color = "progress-bar-danger";
break;
case $number == range(3 , 5):
$color = "progress-bar-warning";
break;
case $number == range(5 , 6):
$color = "progress-bar-default";
break;
case $number == range(6 , 8):
$color = "progress-bar-success";
break;
case $number == range(8, 10):
$color = "progress-bar-success";
break;
}
return $color;
}
But in action this function does not work for me. How should I fix this ?
You are comparing range() which is an array, and $number is integer, which is invalid,
Change your function something like,
function color_switch($number) {
switch ($number) { // switching the function argument
case $number <= 3 : // if less than three, execute case
$color = "progress-bar-danger";
break;
case $number <= 5 :
$color = "progress-bar-warning";
break;
case $number <= 6 :
$color = "progress-bar-default";
break;
case $number <= 8 :
$color = "progress-bar-success";
break;
case $number <= 10 :
$color = "progress-bar-success";
break;
}
return $color;
}
Your utilization of switch is incorrect and your utilization of range() is too.
Your parameter of switch should be the variable you evaluate.
Range() will return an array containing the range.
So the correct code is better :
function color_switch($number) {
switch ($number){
case in_array($number, range(1 , 3)):
$color = "progress-bar-danger";
break;
case in_array($number, range(3 , 5)):
$color = "progress-bar-warning";
break;
case in_array($number, range(5 , 6)):
$color = "progress-bar-default";
break;
case in_array($number, range(6 , 8)):
$color = "progress-bar-success";
break;
case in_array($number, range(8 , 10)):
$color = "progress-bar-success";
break;
}
return $color;
}
I Want to make a calculator. I made the code but problem is when some number divide by zero it gives an Exception,so it is unable to generate the correct result, is there a fix for this? see code
<?php
$x = $_POST['x'];
$y = $_POST['y'];
$tafel = 10;
$antwoord = '' ;
{
if($y = 0)
return 'Voer voor y een ander getal in';
}
switch ($_POST['type']) {
case 'plus':
$antwoord = $x + $y;
break;
case 'keer':
$antwoord = $x * $y;
break;
case 'wortel':
$antwoord = sqrt($x);
break;
case 'min':
$antwoord = $x - $y;
break;
case 'deel':
$antwoord = $x / $y;
break;
case 'kwadraat':
$antwoord = pow($x, 2);
break;
case 'macht':
$antwoord = pow($x, $y);
break;
case 'tafel':
for($i = $x; $i <= 10; $i++){
echo $x * $i.'<br>';
}
break;
default:
# code...
break;
}
echo '<h1 id="answer"> Antwoord:</h1><br>';
echo $antwoord;
?>
The following statement will always be true as you're actually assigning the value within your if statement, instead of comparing it.
if($y = 0)
to
if(0 == $y)
Just change this:
(If this code is not in a function)
return 'Voer voor y een ander getal in';
to this:
die('Voer voor y een ander getal in');
So that the script stops! And you have to make a comparison like this:
if($y == 0)
//^^ See here 2x '='
For more information about comparison operator see the manual: http://php.net/manual/en/language.operators.comparison.php
The following code work fine for you
= means assign
== means comparison
in your situation your need is to compare in if condition
<?php
$x = $_POST['x'];
$y = $_POST['y'];
$tafel = 10;
$antwoord = '' ;
{
if($y == 0)
return 'Voer voor y een ander getal in';
}
switch ($_POST['type']) {
case 'plus':
$antwoord = $x + $y;
break;
case 'keer':
$antwoord = $x * $y;
break;
case 'wortel':
$antwoord = sqrt($x);
break;
case 'min':
$antwoord = $x - $y;
break;
case 'deel':
$antwoord = $x / $y;
break;
case 'kwadraat':
$antwoord = pow($x, 2);
break;
case 'macht':
$antwoord = pow($x, $y);
break;
case 'tafel':
for($i = $x; $i <= 10; $i++){
echo $x * $i.'<br>';
}
break;
default:
# code...
break;
}
echo '<h1 id="answer"> Antwoord:</h1><br>';
echo $antwoord;
?>
I am having a small problem with my PHP MySQL Select. The function is inside of a PHP class. Here is the error I get:
Warning: mysql_fetch_array() expects parameter 1 to be resource,
integer given in C:\xampp\htdocs\include\database.php on line 59
Warning: extract() expects parameter 1 to be array, null given in
C:\xampp\htdocs\include\database.php on line 59
The function just simply updates the database to show what browser and OS they visited the site with. The function is called from another file that is called by an AJAX call that uses POST to send the data about the OS and browser that was gathered from a Javascript file. It only fails if there is an entry of the IP address already in the database. If there is no IP Address entry in the database it succeeds in creating one.
Here is my code:
function addStat($browser, $os){
$IE = 0; $Firefox = 0; $Safari = 0; $Opera = 0; $Chrome = 0; $otherb = 0;
$Windows = 0; $Linux = 0; $Mac = 0; $Android = 0; $iOS = 0; $otheros = 0;
$ql = 0; $totalVisits = 0;
$ip = ip2long($_SERVER['REMOTE_ADDR']);
$q1 = mysql_query("SELECT * FROM " . DB_STATS . " WHERE ip='$ip'", $this->connection);
if (mysql_num_rows($q1)==0){
$browser = mysql_real_escape_string($browser);
$os = mysql_real_escape_string($os);
switch($browser){
case "Internet Explorer":
$IE += 1;
break;
case "Firefox":
$Firefox += 1;
break;
case "Safari":
$Safari += 1;
break;
case "Opera":
$Opera += 1;
break;
case "Chrome":
$Chrome += 1;
break;
default:
$otherb += 1;
break;
}
switch($os){
case "Windows":
$Windows += 1;
break;
case "Mac OS X":
$Mac += 1;
break;
case "Linux":
$Linux += 1;
break;
case "Android":
$Android += 1;
break;
case "iOS":
$iOS += 1;
break;
default:
$otheros += 1;
break;
}
$q = $this->query("INSERT INTO " . DB_STATS . " VALUES (null, '$ip', '$Chrome', '$IE', '$Firefox', '$Opera', '$Safari', '$otherb', '$Windows', '$Mac', '$Linux', '$Android' , '$iOS' , '$otheros', 1)");
if ($q == true){
return(1);
}
else{
return(0);
}
}
else if (mysql_num_rows($q1)==1){
extract(mysql_fetch_array($ql));
switch($browser){
case "Internet Explorer":
$IE += 1;
break;
case "Firefox":
$Firefox += 1;
break;
case "Safari":
$Safari += 1;
break;
case "Opera":
$Opera += 1;
break;
case "Chrome":
$Chrome += 1;
break;
default:
$otherb += 1;
break;
}
switch($os){
case "Windows":
$Windows += 1;
break;
case "Mac OS X":
$Mac += 1;
break;
case "Linux":
$Linux += 1;
break;
case "Android":
$Android += 1;
break;
case "iOS":
$iOS += 1;
break;
default:
$otheros += 1;
break;
}
$totalVisits += 1;
$q = $this->query("UPDATE " . DB_STATS . " set Chrome='$Chrome', IE='$IE', Firefox='$Firefox', Opera='$Opera', Safari='$Safari', otherb='$otherb', Windows='$Windows', Mac='$Mac', Linux='$Linux', Android='$Android' , iOS='$iOS' , otheros='$otheros', totalVisits='$totalVisits'");
if ($q == true){
return(1);
}
else{
return(0);
}
}
else{
return(-1);
}
}
I hope everything made sense and that someone will help.
I see it now -- you used $ql (lower case L) when you intend to use $q1. Let this be a lesson against using very short variable names or very similar names.
// $ql was initialized to 0
$ql = 0; $totalVisits = 0;
// $q1 holds the result resource
extract(mysql_fetch_array($q1));
It is not advisable to call extract() on the output of mysql_fetch_array() unless you also specify the second parameter MYSQL_ASSOC as the fetch type. By default it returns both numeric and associative indices for each column.
extract(mysql_fetch_array($q1, MYSQL_ASSOC));
// Or better
extract(mysql_fetch_assoc($q1));
In general, I would probably advise against using extract() in most any situation, since it results in numerous variables dumped into the global namespace, in particular when you have done SELECT * without being specific about which columns are selected. Better to access them via their array:
$row = mysql_fetch_assoc($q1);
echo $row['browser'];