Hello I have a foreach loop that assigns values to the following code
${'varimages' . $image_count} = $images;
// outputs for image_count = 1, $varimages1 = a string
However, I want to make an IF statement to check something. The "correct" form should be below but I have errors for the presence of "{" and "}"
if (${'varimages' . $image_count} == ${'vartitle' . $image_count})
{ echo "ok"; } else { echo "not ok"; }
How can I rewrite that statement correctly? Thank u!
Using arrays, it would be:
if($varimages[$image_count] == $vartitle[$image_count])
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 $subfieldText2;
Im writing a page in HTML/PHP that connects to a Marina Database(boats,owners etc...) that takes a boat name chosen from a drop down list and then displays all the service that boat has had done on it.
here is my relevant code...
if(isset($_POST['form1'])){//if there was input data submitted
$form1 = $_POST['form1'];
$sql1 = 'select Status from ServiceRequest,MarinaSlip where MarinaSlip.SlipID = ServiceRequest.SlipID and BoatName = "'.$form1.'"';
$form1 = null;
$result1 = $conn->query($sql1);
$test = 0;
while ($row = mysqli_fetch_array($result1, MYSQLI_ASSOC)) {
$values1[] = array(
'Status' => $row['Status']
);
$test = 1;
}
echo '<p>Service Done:</p><ol>';
if($test = 1){
foreach($values1 as $v1){
echo '<li>'.$v1['Status'].'</li>';
}
echo '</ol>';
}else{
echo 'No service Done';
}
the issue im having is that some of the descriptions of sevice are simply Open which i do not want displayed as service done, or there is no service completed at all, which throws undefined variable: values1
how would I stop my script from adding Open to the values1 array and display a message that no work has been completed if values1 is empty?
Try this
$arr = array();
if (empty($arr))
{
echo'empty array';
}
We often use empty($array_name) to check whether it is empty or not
<?php
if(!empty($array_name))
{
//not empty
}
else
{
//empty
}
there is also another way we can double sure about is using count() function
if(count($array_name) > 0)
{
//not empty
}
else
{
//empty
}
?>
To make sure an array is empty you can use count() and empty() both. but count() is slightly slower than empty().count() returns the number of element present in an array.
$arr=array();
if(count($arr)==0){
//your code here
}
try this
if(isset($array_name) && !empty($array_name))
{
//not empty
}
You can try this-
if (empty($somelist)) {
// list is empty.
}
I often use empty（$arr) to do it.
Try this instead:
if (!$values1) {
echo "No work has been completed";
} else {
//Do staffs here
}
I think what you need is to check if $values1 exists so try using isset() to do that and there is no need to use the $test var:
if(isset($values1))
foreach($values1 as $v1){
echo '<li>'.$v1['Status'].'</li>';
}
Or try to define $values1 before the while:
$values1 = array();
then check if it's not empty:
if($values1 != '')
foreach($values1 as $v1){
echo '<li>'.$v1['Status'].'</li>';
}
All you have to do is get the boolean value of
empty($array). It will return false if the array is empty.
You could use empty($varName) for multiple uses.
For more reference : http://php.net/manual/en/function.empty.php
This question already has answers here:
The 3 different equals
(5 answers)
Closed 8 years ago.
<?php
if ($user->getProfile()->get('title')="Canon"); {
echo "Test1"; }
else {
echo "Test2"; }
?>
This is causing my site to break, is there an obvious mistake?
Thank you.
= is an assignment, you want == for a comparison.
You also shouldn't have a ; between ) and {.
This will work:
<?php
if ( ($user->getProfile()->get('title')) == "Canon" ){
echo "Test1";
}
else {
echo "Test2";
}
?>
First of all You need a comparison '==' and not an assigning '='
And then You have a syntax error with a ';' after the condition
You need to change the = sign to == as the first one assigns a values while the latter one compares it.
And also, you don't need to terminate the if statement with a semicolon
if ($user->getProfile()->get('title') == "Canon") /* note that there is no semicolon here */
{ echo "Test1"; }
else
{ echo "Test2"; }
<?php
if ($user->getProfile()->get('title')=="Canon") {
echo "Test1";
}
else {
echo "Test2";
}
?>
Try this.
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 check if the mysql_fetch_array() function returns an empty array or not. But my code doesn't seem to work. Here I want to ensure that if the array is empty I want to display under construction message.
Code :
$queryContents= queryMembers();
$exeQuery = mysql_query($queryContents);
while($fetchSet = mysql_fetch_array($exeQuery)) {
if(count($fetchSet) == 0) {
echo "This Page is Under Construction";
}else{
// something else to display the content
}
}
How do I check to acheive such feature ?
use mysql_num_rows to count number of rows. try this.
$exeQuery = mysql_query($queryContents);
if(mysql_num_rows($exeQuery)== 0){
echo "This Page is Under Construction";
}
else{
while($fetchSet = mysql_fetch_array($exeQuery)) {
// something else to display the content
}
}
You really should be using mysql_num_rows http://us2.php.net/manual/en/function.mysql-num-rows.php
However, on a side note, you should use php empty() instead. http://us2.php.net/empty
When you use mysql_fetch_array(), it returns the rows from the data
set one by one as you use the while loop.
If there will be no record, while loop wont execute. In this case, declare a boolean variable and make it true if it enters the while loop. Like:
$queryContents= queryMembers();
$exeQuery = mysql_query($queryContents);
$recordExists = 0;
while($fetchSet = mysql_fetch_array($exeQuery)) {
if($recordExists == 0 )
$recordExists = 1;
// something else to display the content
}
if($recordExists == 0 ){
echo "This Page is Under Construction";
}
Hope this works!
You can do it this way:
while($r[]=mysql_fetch_array($sql));
// now $r has all the results
if(empty($r)){
// do something
}
source: php doc
Your code inside the while loop never runs if there are no results. mysql_fetch_array returns null/false if there are no more results. What you need yo do is check with mysql_num_rows first, before the while.
$queryContents= queryMembers();
$exeQuery = mysql_query($queryContents);
if(mysql_num_rows ($exeQuery) == 0) {
echo "This Page is Under Construction";
}
while($fetchSet = mysql_fetch_array($exeQuery)) {
// something else to display the content
}
Try this
if(empty($fetchSet)
{
echo "This Page is Under Construction";
}
else
{
// something else to display the content
}