So i have a problem getting all of the results from my while loop within a function.When i print the results of the function its just 1 result instead of 5-10 which i need.There is the code:
function display_post_edits($post_id,$post_name){
global $connect;
$result = array();
$rev = sanitize('revision');
$post_name = '$post_id-revision-v1'; //sanitize this soon!!
$query = mysqli_query($connect,"SELECT post_author,post_date FROM posts WHERE post_type ='$rev' AND post_name = '$post_id-revision-v1'");
while ($row = mysqli_fetch_assoc($query)){
$result[]= $row;
}
return $result;
}
//outside of function
$display_edits = display_post_edits(get_post_id($_POST['subject']),$_POST['subject']);
print_r($display_edits);
So i am trying to get all of the results from the while loop and attach it to $display_edits and when i need for example post date only i will use $display_edits['post_date'].
There are some of my tries so far:
foreach($result as $row){
return implode($result[0]);
}
Returns 1 results because $return is printing arrays like http://prntscr.com/gby0xu (array in array) so i need to define a index.Same thing with implode i need to define index.
while ($row = mysqli_fetch_assoc($query)){
$result = array(
'post_author' =>$row'post_author',
'post_date' =>$row'post_date'
);
}
Prints only 1 results again...
If you need anything more,or i missed something to give here please say what and i will add it !
Thanks in advance !
You foreach should be like this:
foreach($result as $row){
return implode($row[0]);
}
It should use $row and not $result
Related
I'm trying to put data from my database into seperate arrays within another array. This works but when I'm trying to fetch the 'user_id' information, it only shows one number so it works like a string. How can I get it to work like an array and get the entire user_id?
$fetch = mysqli_query($con, "SELECT * FROM spotify_userdata");
$return_arr = [];
while ($row = mysqli_fetch_array($fetch, MYSQL_ASSOC)) {
$return_arr[] = array(
$row_array['user_id'] = $row['user_id'],
$row_array['name'] = $row['name'],
$row_array['artists'] = $row['artists'],
);
}
$user = json_encode($return_arr[0]);
echo $user[2];
This code returns 1 so it show the third number of the user_id. How can I get it to show the entire user_id like this: 111434343
You have many things in your code that's wrong:
Remove the last array item's comma
Change
$return_arr = [];
To
$return_arr = array();
3.Add:
$row_array = array()
at the begginning of all that code
At the end your code must be like this:
$fetch = mysqli_query($con, "SELECT * FROM spotify_userdata");
$row_array = array();
while ($row = mysqli_fetch_array($fetch, MYSQL_ASSOC)) {
$return_arr = array(
$row_array['user_id'] = $row['user_id'],
$row_array['name'] = $row['name'],
$row_array['artists'] = $row['artists'],
);
}
$user = json_encode($return_arr[0]);
According to your code you should use:
echo $user['user_id'];
But the real problem is - where is $row_array initialized?!?
And even bigger problem - why use "=" inside array creation in the while ... it seems to me that "=>" would fit better, don't you think?
I have this PHP function i am using to retrieve rows from a mysql database:
$stmt = $pdo_conn->prepare("SELECT * from admin where sequence > :sequence ");
$stmt->execute(array(':sequence' => $user_sequence));
$records = $stmt->fetchAll(PDO::FETCH_ASSOC);
$results=array();
foreach($records as $results) {
return $results;
}
here i am calling the function:
$test = AdminUserSessionID2('2');
echo $test["forename"];
but it is only displaying one row, what have i done wrong which is making it not display all rows?
Why return in foreach? Of course it will return just the first row. It's like saying foreach(rows as row){ return row; }.
<?php
function MyFunction($user_sequence){
global $pdo_conn;
$stmt = $pdo_conn->prepare("SELECT * from admin where sequence > :sequence;");
$stmt->execute(array(':sequence' => $user_sequence));
$records = $stmt->fetchAll(PDO::FETCH_ASSOC);
return $records;
}
var_dump(MyFunction($user_sequence));
?>
You are assigning results as an empty array.
Then assigning it as the first item of records and returning it.
Try:
foreach($records as $row) {
array_push($results, $row)
}
You can't return multiple data/results in a foreach loop, because the first return will end the function. It's better to return the full array/records and do a foreach loop outside the function.
The $results array is getting returned, which should then be looped outside of the function
Try this:
function AdminUserSessionID2($user_sequence){
$stmt = $pdo_conn->prepare('SELECT * from admin where sequence > :sequence');
$stmt->execute(array(':sequence' => $user_sequence));
$records = $stmt->fetchAll(PDO::FETCH_ASSOC);
return $records;
}
No-one has pointed out that return will exit your current function directly preventing any further processing, and this is why you get just one row output. Your specific code based on what you want to achieve would be:
$stmt = $pdo_conn->prepare("SELECT * from admin where sequence > :sequence ");
$stmt->execute(array(':sequence' => $user_sequence));
$records = $stmt->fetchAll(PDO::FETCH_ASSOC);
$results=array();
foreach($records as $results) {
$test = AdminUserSessionID2('2');
echo $test["forename"];
}
See: http://php.net/return
This might seem like a really easy question but it has got me stumped lol. I am trying to print the rows received from the database. I want to store the rows inside an array and then print them using a for loop. I know that the query works however when I try to print the array elements it only prints the word array. I have tired doing it with a foreach loop and a simple for loop. If anyone can point me in the right direction would be a life saver.
Printing Php Code
<?php
$type = "FREE";
$free = getTerms($type);
echo "<p>";
for($j = 0; $j < count($free); $j++)
{
echo "start".$free[$j]."end";
}
echo "</p>";
?>
geting the rows from the database
function getTerms($type)
{
$terms = array();
$connection = mysql_open();
$query = "select terms from terms_and_con where accountType='$type' && currentTerms='YES'";
$results = mysql_query($query, $connection) or show_error("signUp.php", "", "");
while($row = mysql_fetch_array($results))
{
$terms[] = $row;
}
mysql_close($connection) or show_error("signUp.php", "", "");
return $terms;
}
Each entry in the $free array is itself an array (from $row).
Try
echo 'start', $free[$j]['terms'], 'end';
Alternatively, you may find a foreach loop more semantically appropriate
foreach ($free as $row) {
echo 'start', $row['terms'], 'end';
}
Edit: I'd advise using mysql_fetch_assoc() instead of mysql_fetch_array() if you're only going to use associative entries from $row.
the thing is function mysql_fetch_array ( as the name suggests) returns an ( in your case both associative and number) array. so $row is actually array(0 => VALUE, 'terms' => 'VALUE')
So what you are trying to echo is actually an array.
Simple fix:
replace:
$terms[] = $row;
with:
$terms[] = $row[0];
It must be Monday, the heat or me being stupid (prob the latter), but for the life of me I cannot get a simple php function to work.
I have a simple query
$sql = mysql_query("SELECT * FROM table WHERE field_name = '$input'");
Which I want to run through a function: say:
function functionname($input){
global $field1;
global $field2;
$sql = mysql_query("SELECT * FROM table WHERE field_name = '$input'");
while($row = mysql_fetch_array($sql)) :
$field1[] = $row['field1'];
$field2[] = $row['field2'];
endwhile;
mysql_free_result($sql);
}
So that I can call the function in numerious places with differeing "inputs". Then loop through the results with a foreach loop.
Works fine the first time the function is called, but always gives errors there after.
As said "It must be Monday, the heat or me being stupid (prob the latter)".
Suggestions please as I really only want 1 function to call rather than rewrite the query each and every time.
This is the error message
Fatal error: [] operator not supported for strings in C:\xampp\htdocs\functions.php on line 270
function functionname($input){
$sql = mysql_query("SELECT field1,field2 FROM table WHERE field_name = '$input'");
$result = array('field1' => array()
'field2' => array()
);
while($row = mysql_fetch_array($sql)) :
$result['field1'][] = $row['field1'];
$result['field2'][] = $row['field2'];
endwhile;
mysql_free_result($sql);
return $result;
}
it seems that somewhere the $field1 or $field2 are converted to strings and you cant apply the [] to a string...
i'd say that you have to do:
$field1 = array();
$field2 = array();
before the WHILE loop
The problem is that you so called arrays are strings!
global $field1;
global $field2;
var_dump($feild1,$feild2); //Will tell you that there strings
Read the error properly !
[] operator not supported for strings
And the only place your using the [] is withing the $feild - X values
GLOBAL must work because the error is telling you a data-type, i.e string so they must have been imported into scope.
another thing, why you selecting all columns when your only using 2 of them, change your query to so:
$sql = mysql_query("SELECT feild1,feild2 FROM table WHERE field_name = '$input'");
another thing is that your using mysql_fetch_array witch returns an integer indexed array, where as you want mysql_fetch_assoc to get the keys.
while($row = mysql_fetch_assoc($sql)) :
$field1[] = $row['field1'];
$field2[] = $row['field2'];
endwhile;
What I would do
function SomeFunction($variable,&$array_a,&$array_b)
{
$sql = mysql_query("SELECT field1,field2 FROM table WHERE field_name = '$variable'");
while($row = mysql_fetch_assoc($sql))
{
$array_a[] = $row['field1'];
$array_b[] = $row['field2'];
}
mysql_free_result($sql);
}
Then use like so.
$a = array();
$b = array();
SomeFunction('Hello World',&$a,&$b);
In my opinion, it's pretty unusual and even useless approach at all.
This function is too localized.
To make a general purpose function would be a way better.
<?
function dbgetarr(){
$a = array();
$query = array_shift($args);
foreach ($args as $key => $val) {
$args[$key] = "'".mysql_real_escape_string($val)."'";
}
$query = vsprintf($query, $args);
$res = mysql_query($query);
if (!$res) {
trigger_error("dbgetarr: ".mysql_error()." in ".$query);
return FALSE;
} else {
while($row = mysql_fetch_assoc($res)) $a[]=$row;
}
return $a;
}
and then call it like this
$data = dbgetarr("SELECT field1,field2 FROM table WHERE field_name = %s",$input);
foreach ($data as $row) {
echo $row['field1']." ".$row['field1']."<br>\n";
}
To understand your issue, we need the error, however, are you sure you are going about this in the right way?
Why do you need to call the function multiple times if you are just changing the value of the input field?
You could improve your SQL statement to return the complete result set that you need the first time.. i.e. SELECT * FROM table GROUP BY field_name;
Not sure if that approach works in your scenario, but in general you should aim to reduce the number of round trips to your database.
I don't know, i right or not. But i advise to try this:
function functionname($input){
global $field1;
global $field2;
$sql = mysql_query("SELECT * FROM `table` WHERE `field_name` = '" . $input . "'");
while($row = mysql_fetch_assoc($sql)) :
$field1[] = $row['field1'];
$field2[] = $row['field2'];
endwhile;
mysql_free_result($sql);
}
I have the following:
while($myRow = odbc_fetch_array( $result )){ <--lots of rows
$thisResult['name'] = $myRow["name"] ;
$thisResult['race'] = $myRow["race"] ;
$thisResult['sex'] = $myRow["sex"];
$thisResult['dob'] = $myRow["dob"];
}
I can't figure out how to print this back out.
I want to get each row and iterate through each row in the array like a datareader. I'm not sure what to do. I do not want to do the echo in the while. I need to be able to print it out elsewhere. But I don't think I've done it right here to be able to print it later.
I also tried, this, however:
while($myRow = odbc_fetch_array( $result )){ <--lots of rows
print($thisResult[$myRow["name"]] = $myRow);
}
I then tried:
while($myRow = odbc_fetch_array( $result )){ <--lots of rows
print (odbc_result($myRow,"name"));
}
but got an error.
Thank you for any help.
EDIT: when I do this:
while($myRow = odbc_fetch_array( $result )){
print ($myRow["name"]);
}
I get undefined index name. I am mainly concerned with saving to an array but I have to be able to do it in the loop first.
Declare an array before and assign the values to it:
$rows = array();
while($myRow = odbc_fetch_array( $result )){ <--lots of rows
$rows[] = $myRow;
}
Then you can print it e.g. this way:
foreach($rows as $row) {
foreach($row as $key => $value) {
echo $key . ': '. $value;
}
}
or however you want to.
You don't have to access and assign $thisResult['name'] = $myRow["name"] in your while loop as $myRow already is an array. You just copy the values which is unnecessary.
You say you have a lot of rows. Depending of what you really want to do with data, it might be better to put all this functionality into the while loop to avoid creating an array.
How about something like:
$output = '';
while($myRow = odbc_fetch_array( $result )) {
$output = $output."Your name is {$myRow["name"]} and your race is {$myRow["race"]}\n";
}
// print output later...