I have two combo boxes, one to report scores and one to set who scored the goals, How can i make it so if on post['submit']
If $_POST['Score1'] and $_POST['Score2']
is other than equal to
$_POST['homegoalscorer1'] and $_POST['awaygoalscorer1']
then echo"fail";
Something like;
if(isset($_POST['submit']))
{
$homescore = $_POST['Score1'];
$awayscore = $_POST['Score2'];
$homegoalscorer = $_POST['homegoalscorer1'];
$awaygoalscorer = $_POST['awaygoalscorer1'];
if '$homescore' + '$awayscore' != $homegoalscorer + $awaygoalscorer {
echo "failed";
}
else {
}
}
Any ideas?
Single quotes on a variable will turn that variable intro a string without execution. Also you forgot to add brackets:
if(isset($_POST['submit'])) {
$homescore = (float)$_POST['Score1'];
$awayscore = (float)$_POST['Score2'];
$homegoalscorer = (float)$_POST['homegoalscorer1'];
$awaygoalscorer = (float)$_POST['awaygoalscorer1'];
if (($homescore+$awayscore) != ($homegoalscorer+$awaygoalscorer)) {
echo "failed";
} else {
}
}
Use some brackets in your if statement to force the conditional setting in the correct context - and why are you encapsulating your variables in single quotes?
if (($homescore + $awayscore) != ($homegoalscorer + $awaygoalscorer))
{
// Your code continues....
Related
I'm trying to get a n else statement working with a print_r such that if there's no value it outputs "no values".
In the code I'm getting values from json converted to an array.
The logic I'm trying to achieve is
IF fieldTag contains "i" THEN output the content associated with it
ELSE says its empty.
Right now blank is outputted as opposed to "no values".
Thanks
for($b=0; $b<count($res['entries'][$i]['bib']['varFields']); $b++) //loop thru the varFields
{
if($res['entries'][$i]['bib']['varFields'][$b]['fieldTag'] == "i")
{
$subfieldText2 = $res['entries'][$i]['bib']['varFields'][$b]['subfields'][0]['content']."<br>";
if(count($subfieldText2) > 0) {
print_r($subfieldText2);
} else {
echo "no values";
}
}
}
count() is for arrays, not strings, the way to get the length of a string is with strlen(). And if you want to check for an empty string, just compare it with $var == "", you don't need to get the length.
But you're concatenating "<br>" to the value, so the length will never be zero. You could check the length before concatenating.
$subfieldText2 = $res['entries'][$i]['bib']['varFields'][$b]['subfields'][0]['content'];
if($subfieldText2 != "") {
$subfieldText2 .= "<br>";
print_r($subfieldText2);
} else {
echo "no values";
}
And to avoid having to repeat that long expression to access the field, you could use foreach
for($res['entries'][$i]['bib']['varFields'] as $field) {
if ($field['fieldTag'] == "i") {
$subfieldText2 = $field['subfields'][0]['content'];
...
}
}
this worked for me thanks everyone
$subfieldText2="not detected";
echo "ISBN: ";
for($b=0; $b<count($res['entries'][$i]['bib']['varFields']); $b++) //loop thru the varFields
{
if($res['entries'][$i]['bib']['varFields'][$b]['fieldTag'] == "i")
{
$subfieldText2 = $res['entries'][$i]['bib']['varFields'][$b]['subfields'][0]['content'];
echo $subfieldText2.", ";
}
}
echo $subfieldText2;
I have a php script for check the availability of some data. I call this script from external jquery. the jquery is running fine. here is my php:
<?php
$avares = checkAva($fi_nm, $tbl_nm, $txtval);
echo $avares;
function checkAva($field, $table, $curval) {
$avres = mysql_query("SELECT " . $field . " FROM " . $table . "") or die("query failed");
while ($a_row = mysql_fetch_array($avres)) {
$dbval = $a_row[$field];
if ($curval == $dbval) {
return "no";
} else {
return "yes";
}
}
}
?>
$curval is the variable coming from external jquery. my problem is that the while loop seems to run only once though there are lot of entries in the DB. I checked it with an integer variable and the while loop seems to run only once. can you help me to solve that?
Look at your code.
while ($a_row = mysql_fetch_array($avres)) {
$dbval = $a_row[$field];
if ($curval == $dbval) {
return "no";
} else {
return "yes";
}
}
you have used return, if its true it returns and false then also returns change those according to your needs. The return statement immediately ends execution of the current function
It will by design as you have a return statement. From what you have said your not actually wanting it to return but to set a variable that at end of execution will return no or yes. I could be wrong on this but hey ho.
<?php
echo checkAva($fi_nm, $tbl_nm, $txtval);
function checkAva($field, $table, $curval) {
$avres = mysql_query("SELECT " . $field . " FROM " . $table) or die("query failed");
$noOrYes = "yes";
while ($a_row = mysql_fetch_array($avres)) {
if($curval == $a_row[$field]) {
$noOrYes = "no";
}
}
return $noOrYes;
}
?>
The possible issue that can cause Loop to iterate once are:
Error in the Variable used for the $query and $result
Same name Variable inside and outside of the Loop
Incorrect placement of Return statement
Invalid Mysql Statement
Directly put the condition in your Query like
function checkAva($field, $table, $curval) {
$avres = mysql_query("SELECT " . $field . " FROM " . $table . "
WHERE `".$field."` = '".$curVal."'");
$res = mysql_fetch_array($avres);
if(is_array($res) && count($res) > 0)
return "Yes";
else
return "No";
}
Instead of getting all the results and checking with each one of the result you directly put a condition to extract the results which satisfies your condition.This will be suggestable if you have many records.
You need to put one of the return outside of the while loop.
For example if you just wanted to check if $curval == $dbval
while ($a_row = mysql_fetch_array($avres)) {
$dbval = $a_row[$field];
//If the condition was met return with a no
if ($curval == $dbval) {
return "no";
}
}
//If the condition was not met return yes
return yes;
That's basically what you need to do so the loop will run until your condition was met or not at all.
I am trying to GET different rows from different columns in php/mysql, and pack them into an array. I am able to successfully GET a jason encoded array back IF all values in the GET string match. However, if there is no match, the code echos 'no match', and without the array. I know this is because of the way my code is formatted. What I would like help figuring out, is how to format my code so that it just displays "null" in the array for the match it couldn't find.
Here is my code:
include '../db/dbcon.php';
$res = $mysqli->query($q1) or trigger_error($mysqli->error."[$q1]");
if ($res) {
if($res->num_rows === 0)
{
echo json_encode($fbaddra);
}
else
{
while($row = $res->fetch_array(MYSQLI_BOTH)) {
if($_GET['a'] == "fbaddra") {
if ($row['facebook'] === $_GET['facebook']) {
$fbaddr = $row['addr'];
} else {
$fbaddr = null;
}
if ($row['facebookp'] === $_GET['facebookp']) {
$fbpaddr = $row['addr'];
} else {
$fbpaddr = null;
}
$fbaddra = (array('facebook' => $fbaddr, 'facebookp' => $fbpaddr));
echo json_encode($fbaddra);
}
}
}
$mysqli->close();
UPDATE: The GET Request
I would like the GET request below to return the full array, with whatever value that didn't match as 'null' inside the array.
domain.com/api/core/engine.php?a=fbaddra&facebook=username&facebookp=pagename
The GET above currently returns null.
Requests that work:
domain.com/api/core/engine.php?a=fbaddra&facebook=username or domain.com/api/core/engine.php?a=fbaddra&facebookp=pagename
These requests return the full array with the values that match, or null for the values that don't.
TL;DR
I need assistance figuring out how to format code to give back the full array with a value of 'null' for no match found in a row.
rather than assigning as 'null' assign null. Your full code as follows :
include '../db/dbcon.php';
$res = $mysqli->query($q1) or trigger_error($mysqli->error."[$q1]");
if ($res) {
if($res->num_rows === 0)
{
echo json_encode('no match');
}
else
{
while($row = $res->fetch_array(MYSQLI_BOTH)) {
if($_GET['a'] == "fbaddra") {
if ($row['facebook'] === $_GET['facebook']) {
$fbaddr = $row['dogeaddr'];
//echo json_encode($row['dogeaddr']);
} else {
$fpaddr = null;
}
if ($row['facebookp'] === $_GET['facebookp']) {
$fbpaddr = $row['dogeaddr'];
//echo json_encode($row['dogeaddr']);
} else {
$fbpaddr = null;
}
$fbaddra = (array('facebook' => $fbaddr, 'facebookp' => $fbpaddr));
echo json_encode($fbaddra);
}
}
}
$mysqli->close();
You can even leave else part altogether.
Check your code in this fragment you not use same names for variables:
if ($row['facebook'] === $_GET['facebook']) {
$fbaddr = $row['dogeaddr'];
//echo json_encode($row['dogeaddr']);
} else {
$fpaddr = 'null';
}
$fbaddr not is same as $fpaddr, this assign wrong result to if statement.
It was the mysql query that was the problem.
For those who come across this, and need something similar, you'll need to format your query like this:
** MYSQL QUERY **
if ($_GET['PUTVALUEHERE']) {
$g = $_GET['PUTVALUEHERE'];
$gq = $mysqli->real_escape_string($g);
$q1 = "SELECT * FROM `addrbook` WHERE `facebookp` = '".$gq."' OR `facebook` = '".$gq."'";
}
** PHP CODE **
if($_GET['PUTVALUEHERE']{
echo json_encode($row['addr']);
}
Good day guys,
I've made a sweet favorites function with php mysql and ajax, and its working great. Now I want to show 'favorite' when favorite = 0 and show 'unfavorite' when favorite = 1
if ($favorites == 0) {
$favorite = 'Favorite';
}
if ($favorites == 1) {
$unfavorite = 'unFavorite';
}
and echo it in the row as :
<div id="favorites">' .($favorite). ' ' .($unfavorite). '</div>
The problem is: when favorite = 0, both $favorite and $unfavorite are being shown. When favorite = 1 only $unfavorite is being shown correctly. Of course it should be $favorite OR $unfavorite. I assume the problem is clear and simple to you, please assist :)
Thanks in advance
It's easier to use just one variable:
$text = ''
if ($favorites == 0) {
$text = 'Favorite';
} else {
$text = 'unFavorite';
}
...
echo $text;
If you want to check $favorite, you are using the wrong variable in your control statement. Also, it is better coding practice to use elseif rather than if for that second if. One more thing: it's easier to manage one resulting variable.
$output = "";
if ($favorite == 0) {
$output = 'Favorite';
}
elseif ($favorite == 1) {
$output = 'unFavorite';
}
...
echo $output; // Or whatever you want to do with your output
Is $favorites an integer?
Anyway try using three equal signs (===) or else instead of the second if:
if ( $favorites === 0 )
{
// ...
}
else // or if ($favorites === 1)
{
// ...
}
You're making a toggle, so you only need one variable:
if(empty($favourites)){
$fav_toggle = 'Favorite';
} else {
$fav_toggle = 'unFavorite';
}
echo $fav_toggle;
Same code is working on me if I assigned $favorites = 0; or $favorites = 1;
You can also use if else
$favorites = 1;
if ($favorites == 0) {
$favorite = 'Favorite';
}
else if ($favorites == 1) {
$unfavorite = 'unFavorite';
}
function add_new($father,$chName) // add new category
{
if($father = "1" ) {
$result = mysql_query("INSERT into stinky_menu (title,nest_under)
VALUES('".$chName."','1')");
}
else {
$result = mysql_query("UPDATE stinky_menu SET title = '$chName' nest_under = '$father'");
}
}
I am getting the value of father from parent page, but its not going to else condition if its not equal to one.
You’re using the assignment operator = rather than the comparison operator ==. So try this:
if ($father == "1") {
// …
} else {
// …
}
That's because you have
if($father = "1")
You need to use "==". "=" is the assignment operator. You are setting $father equal to "1" even when it isn't.
Try:
if ($father == 1){}
Read here about comparison operators. "=" is the assignment operator.
Look at this to see what your code does:
<?php
$father = 55;
if ($father = 1){}
else{}
echo $father;
?>
This prints "1".
Also, should not that last query be:
"UPDATE stinky_menu SET title = '$chName', nest_under = '$father'"