I am implementing keyword search in PHP. When no. of results from DB are greater than 264 PHP echo fails to give any output. Any idea what might be the problem here?
// my algo is -
$result = mysql_query(searchQueryString) or die(mysql_error());
$row = mysql_fetch_assoc($result);
//Create a $json_array of required values
$str_to_print = json_encode($json_array);
echo $str_to_print;//this gives empty output when no. of rows from DB > 264
//Everything else is working properlycode here
Please excuse this, I wanted to comment but am unable to due to rep.
I used this statement: print json_encode(array(1, 2, 3, 4, 5)); and got the result of: [1,2,3,4,5].
Either your result from the mysql_query is wrong, or something else is.
Just noticed, you are using this line of code:
$str_to_print = json_encode($json_array);
But where is $json_array set? Is it set at all or are you passing a null pointer to the function? Error: Notice: Undefined variable: json_array
Related
i'm still new towards php but after searching through multiple topics i can't seem to figure this one out.
$query2 = "SELECT COUNT(MAP_CODE2) FROM TCODE_MAPPING WHERE MAP_CODE2 = 'ABC123'";
$result2 = odbc_exec($connect,$query2);
echo $result2;
i have a query above that i'd like to use to get the total number of rows within the query that i've set, however for some reason i keep getting hit by the error
Array to string conversion in /var/www/html/xxx.php on line 85
Would highly appreciate if anyone could help out on what i'm doing wrong. Thank you!
Problem is:-
You are trying to echo a result-set object of array type.
Solution (check the code comments):-
$query2 = "SELECT COUNT(MAP_CODE2) AS MYCOUNT FROM TCODE_MAPPING WHERE MAP_CODE2 = 'ABC123'"; // given a name to the count
$result2 = odbc_exec($connect,$query2); //prepare and execute query
while (odbc_fetch_row($result2)) { //iterate over result-set object
echo odbc_result($result, "MYCOUNT"), "\n"; // echo count
}
What is happening here is you are echoing an array object type that is also a resource type and php echo only string and other primitive type variables.
so you need to use a loop to access the row or to get row count and access row just use a foreach($results as $result)
to get row count visit this sqlsrv-num-rows . on this official php link you can get more example how to fetch data.
what in the following code is causeing Warning: Illegal string offset 'quantity' in ' because of PHP 5.4 version cant find a fix?
<?php
include "data.php";
function checkStock($id_film){
$query_stock = "SELECT stockquantity FROM table WHERE id = {$id_film}";
$result_stock = mysql_query($query_stock);
$row = mysql_fetch_row($result_stock);
echo $row[0]['quantity']; //this where the error is
}
If you'd like to debug, you could write:
$row = mysql_fetch_row($result_stock);
echo "<pre>";
print_r($row);
echo "</pre>";
That will give you an idea about what the array looks like. From your MYSQL query, it's probably going to be like:
Array(
"quantity" => 5
)
So, this should work:
echo $row['quantity'];
instead of
echo $row[0]['quantity'];
Hope this helps.
Peace! xD
Obligatory chastisement: Don't use the mysql API; it's deprecated and has been removed completely from PHP 7. Now that that's out of the way...
$row = mysql_fetch_row($result_stock); returns a single row of data as a one-dimensional array. Your reference to $row[0]['quantity'] is attempting to dereference a two-dimensional array, so of course you're getting an error because the second dimension doesn't exist.
rjdown is correct; you just want $row['quantity'] which selects the quantity column of the current row.
You have used mysql_fetch_row which returns single row of database as an array which you can access information like $row[0], $row[1], ... In your case it seems that you are accessing one field of a unique record then you have only one field to be returned if found or null if not found. If you want to access information like an associated array use mysql_fetch_assoc then the result will be stored in an associated array like $row['quantity'].
So try to edit this part
$row = mysql_fetch_row($result_stock); echo $row[0]['quantity'];
Like this
$row = mysql_fetch_assoc($result_stock); echo $row['quantity'];
i'm using the following to count the number of rows in my table:
$book_count = query("SELECT status FROM scifi_book WHERE read = $uid");
(count($book_count));
echo $book_count;
and i get the following error:
Notice: Array to string conversion on line 167
(which is the line echo $book_count;)
FIY, the query function is defined and works fine. Never had any issues to insert, select, update and delete.
Am I missing something here?
Try this:
$book_count = query("SELECT status FROM scifi_book WHERE read = $uid");
echo count($book_count);
Also, you need to use print_r($book_count) since your $book_count is not a string.
Suggestion: if you only use that query for getting the count, this may be a bit better:
$book_count = query("SELECT count(*) FROM scifi_book WHERE read = $uid");
your query function seems to return an array, not a string. Instead of echo $follower_count use print_r($follower_count) to see what's inside the query response.
The reason you are seeing that error is because you are echoing Array which is the returned by your query function. echo construct only works on strings, please see the documentation here: http://www.php.net/manual/en/function.echo.php.
Had you used print_r or var_dump then you wouldn't have seen that error. So as #A.S. Roma and Nathaniel Granor suggest use print_r
$book_count = query("SELECT status FROM scifi_book WHERE read =".$uid);
(count($book_count));
echo $book_count;
I am getting this PHP error, what does it mean?
Notice: Undefined offset: 0 in
C:\xampp\htdocs\mywebsite\reddit_vote_tut\src\votes.php on line 41
From this code:
<?php
include("config.php");
function getAllVotes($id)
{
$votes = array();
$q = "SELECT * FROM entries WHERE id = $id";
$r = mysql_query($q);
if(mysql_num_rows($r)==1)//id found in the table
{
$row = mysql_fetch_assoc($r);
$votes[0] = $row['votes_up'];
$votes[1] = $row['votes_down'];
}
return $votes;
}
function getEffectiveVotes($id)
{
$votes = getAllVotes($id);
$effectiveVote = $votes[0] - $votes[1]; //ERROR THROWN HERE
return $effectiveVote;
}
$id = $_POST['id'];
$action = $_POST['action'];
//get the current votes
$cur_votes = getAllVotes($id);
//ok, now update the votes
if($action=='vote_up') //voting up
{
$votes_up = $cur_votes[0]+1; //AND ERROR THROWN HERE
$q = "UPDATE threads SET votes_up = $votes_up WHERE id = $id";
}
elseif($action=='vote_down')
{
$votes_down = $cur_votes[1]+1;
$q = "UPDATE threads SET votes_down = $votes_down WHERE id = $id";
}
$r = mysql_query($q);
if($r)
{
$effectiveVote = getEffectiveVotes($id);
echo $effectiveVote." votes";
}
elseif(!$r) //voting failed
{
echo "Failed!";
}
?>
You are asking for the value at key 0 of $votes. It is an array that does not contain that key.
The array $votes is not set, so when PHP is trying to access the key 0 of the array, it encounters an undefined offset for [0] and [1] and throws the error.
If you have an array:
$votes = array('1','2','3');
We can now access:
$votes[0];
$votes[1];
$votes[2];
If we try and access:
$votes[3];
We will get the error "Notice: Undefined offset: 3"
first, check that the array actually exists, you could try something like
if (isset($votes)) {
// Do bad things to the votes array
}
This answer helped me https://stackoverflow.com/a/18880670/1821607
The reason of crush — index 0 wasn't set. Simple $array = $array + array(null) did the trick. Or you should check whether array element on index 0 is set via isset($array[0]). The second variant is the best approach for me.
Use print_r($votes); to inspect the array $votes, you will see that key 0 does not exist there. It will return NULL and throw that error.
getAllVotes() isn't returning an array with the indexes 0 or 1. Make sure it's returning the data you want by calling var_dump() on the result.
I encountered this as well and the solution is simple, dont hardcode the array index position in your code.
Instead of
$data[0]['somekey'] do
foreach($data as $data_item)
{
echo $data_item['somekey'];
} If there is an array or more you can perform your desired action inside the loop, but if it's undefined it will not bring an error. you can also add other checks on top of that.
You could also add a variable and increment it in a for in loop to limit your looping if you want only the first positions or something.
As explained it happens because there is no data in the $cur_votes[0] and hence it throws an error.
To ensure your code works fine, before performing "$votes_up = $cur_votes[0]+1;" echo the $cur_votes[0] value to see if there is any value stored or not.
Surely, there is no value stored.
function getEffectiveVotes($id)
According to the function header, there is only one parameter variable ($id).
Thus, on line 27, the votes[] array is undefined and out of scope. You need to add another
parameter value to the function header so that function getEffectiveVotes() knows to expect two parameters. I'm rusty, but something like this would work.
function getEffectiveVotes($id, $votes)
I'm not saying this is how it should be done, but you might want to research how PHP
passes its arrays and decide if you need to explicitly state to pass it by reference
function getEffectiveVotes($id &$votes) <---I forget, no time to look it up right now.
Lastly, call function getEffectiveVotes() with both arguments wherever it is supposed to be called.
Cheers.
As you might have already about knew the error. This is due to trying to access the empty array or trying to access the value of empty key of array. In my project, I am dealing with this error with counting the array and displaying result.
You can do it like this:
if(count($votes) == '0'){
echo 'Sorry, no votes are available at the moment.';
}
else{
//do the stuff with votes
}
count($votes) counts the $votes array. If it is equal to zero (0), you can display your custom message or redirect to certain page else you can do stuff with $votes. In this way you can remove the Notice: Undefined offset: 0 in notice in PHP.
If you leave out the brackets then PHP will assign the keys by default.
Try this:
$votes = $row['votes_up'];
$votes = $row['votes_down'];
In my case it was a simple type
$_SESSION['role' == 'ge']
I was missing the correct closing bracket
$_SESSION['role'] == 'ge'
If you are using dompdf/dompdf and error occure in vendor/dompdf/dompdf/src/Cellmap.php then
It looks like we're using the wrong frame id in the update_row_group method. Initial testing seems to confirm this. Though that may be because this is strictly a paged table issue and not too many of the documents in my test bed have paged tables.
Can you try changing line 800 to:
$r_rows = $this->_frames[$g_key]["rows"];
($g_key instead of $r_key)
https://github.com/dompdf/dompdf/issues/1295
Use mysql row instead
mysql_fetch_row($r)
Meanwhile consider using mysqli or PDO
?>
Try seeding data with command: php artisan db:seed.
its just a warning use:
error_reporting(0);
it shows when we do not initialize array and direct assigning value to indexes.
somefunction{
$raja[0]="this";
$raja[1]="that";
}
instead :
somefunction{
$raja=array(0=>'this',1='that');
//or
$raja=array("this","that");
}
it just notification, do not generate any output error or any unexpected output.
I am having great trouble trying to use mysql_fetch_assoc and mysql_result together in the same PHP script.
Originally (when not utilizing the mysql_result function) I was getting the required database values using a combination of mysql_query and mysql_fetch_assoc and everything was fine. Then i added 2 lines into my code to obtain certain ‘title’ field values using mysql_result.
Now if i run my script as it is below i will only receive 1 result even though there are 2 result. Then if i move my do/while loop up so that it is between the other 2 blocks of code (mysql_fetch_assoc and mysql_result lines) i will receive the desired 2 results.
I need my loop to come after the mysql_result section so putting the loop before it is not an option.
// connect to DB and get values
mysql_select_db($database, $mywebsite);
$query_not_related_before = "SELECT * FROM table limit 2";
$not_related_before = mysql_query($query_not_related_before, $ mywebsite);
$row_not_related_before = mysql_fetch_assoc($not_related_before);
// Extract just the results from the title field (the problem area!)
$before_essayid4 = mysql_result($not_related_before,0, 'title');
$before_essayid5 = mysql_result($not_related_before,1, 'title');
// Display results etc
do {
echo "<br />".$row_not_related_before['title']."<br />";}
while ($row_not_related_before = mysql_fetch_assoc($not_related_before));
Plase help,
Many thanks,
David
I am unsure if this will solve your problem but I think you should "seek" the result back.
mysql_data_seek ($not_related_before, 0)
Also, check out the warning on the mysql_result page:
Calls to mysql_result() should not be
mixed with calls to other functions
that deal with the result set.
Hope this helps ;)
You get one row because the first already requested here:
$row_not_related_before = mysql_fetch_assoc($not_related_before);
So I think you have to move result pointer to beginning:
mysql_data_seek($not_related_before, 0);
// Display results etc
do {
echo "<br />".$row_not_related_before['title']."<br />";}
while ($row_not_related_before = mysql_fetch_assoc($not_related_before));
A simpler solution would be to use 2D array's, i have always found the mysql functions to be a little cumbersome.
<?php
$result = mysqli_query(db_connect(),$query);
$result_out = array();
if(#mysqli_num_rows($result))
while($row=mysqli_fetch_array($result,MYSQLI_ASSOC))$result_out[]=$row;
foreach($result_out as $row)
{
echo "<br />".$row['title']."<br />";
}
?>
hth