Currently I fetch rows with the help of mysql_fetch_array() (and the LIMIT/OFFSET) (ie. while ($row = mysql_fetch_array($res)) { ... }); but the last row is not needed as it was to be fetched in order to make sure there is at least one more row left - it is not for display.
You can simply get the number of rows before this loop:
select count(*) from <your_table> WHERE <your_condition>;
Then, you can exclude the last record within the loop using a counter.
Is this what you want??
You should not mix presentation with logic anyway. Fetch all the rows in an array:
$rows = array();
while ($row = mysql_fetch_array($res)) {
$rows[] = $row;
}
and delete the last one:
array_pop($rows);
Or just set LIMIT to get one less of that is possible.
Update:
Here is another possibility using a for loop and mysql_num_rows:
for($i = mysql_num_rows($res) - 1; $i--; $row = mysql_fetch_array($res))
but the first one is more readable imo.
Related
I need to select multiple comments (if there are any) based on the photo_id. As I understand it you can use the WHERE clause but I'm not exactly sure how to select multiple ones and store them in some kind of array?
e.g.
$result = mysqli_query($conn,"SELECT * FROM comments WHERE photo_id='$photo1id'");
$row = $result->fetch_assoc(); // but there's more than 1 row
If for example $photo1id == 21, how do I get all the comments (2 in this case)? Some kind of while loop?
At the end of the PHP file I have this:
echo json_encode(array('photo1id'=>$photo1id));
I need to store each row in that array somehow because I need to retrieve the data in another PHP file using $.getJSON. Or perhaps there is a better solution to this.
Loop through it and generate an array -
while($row = $result->fetch_assoc()) {
$comments[] = $row;
}
After that you can send the array as json.
echo json_encode($comments);
Is there is more rows, you need to use a loop.
while ($row = $result->fetch_assoc()) {
// your code here
}
Try the code below:
//Run query
$result = mysqli_query($conn,"SELECT * FROM comments WHERE photo_id='$photo1id'");
//While there is a result, fetch it
while($row = $result->fetch_assoc()) {
//Do what you need to do with the comment
}
If you don't want to print the code straight away you can just create an array:
$x=0;
while($row = $result->fetch_assoc()) {
$comment[$x]=$row['comment'];
$x++;
}
I can't seem to figure out how this loop works in PHP:
$people = array();
while($row = $result->fetch_assoc())
$people[] = $row;
It seems as though the loop would just keep going, infinitely. But, it doesn't How exactly does this work? Can someone explain it to me step-by-step? I'm guessing that the while loop could also be written like this:
while(($row = $result->fetch_assoc()) == true)
The fetch_assoc fetches one result row at a time and stores it in $row. Since this is in a loop, you are fetching until you run out of rows
In the loop you are essentially pushing the $row value into a $people array
Your code:
$people = array();
while($row = $result->fetch_assoc())
$people[] = $row;
Example how its works (MySQL):
$people = array();
$result = mysql_query($query);
$rows_count = mysql_num_rows($result);
for ($i = $rows_count - 1; $i >= 0; $i--) {
mysql_data_seek($result, $i);
$people[] = mysql_fetch_assoc($result);
}
How U see first option is more compact.
fetch_assoc will return false either upon an error or when the cursor for the fetch hits the end of all the rows and no more rows can be fetched, so it will return false.
Every time you fetch for the query, a cursor keeps track of the last returned row and will continue to keep track ultimately till you finish reading all rows.
Edit: Sorry I was thinking about PDO and not mysqli, however it should be about the same thing.
I'm probably missing something easy, but I seem to be blocked here... I have a MySQL database with two tables and each table has several rows. So the goal is to query the database and display the results in a table, so I start like so:
$query = "SELECT name, email, phone FROM users";
Then I have this PHP code:
$result = mysql_query($query);
Then, I use this to get array:
$row = mysql_fetch_array($result);
At this point, I thought I could simply loop through the $row array and display results in a table. I already have a function to do the looping and displaying of the table, but unfortunately the array seems to be incomplete before it even gets to the function.
To troubleshoot this I use this:
for ($i = 0; $i < count($row); $i++) {
echo $row[$i] . " ";
}
At this point, I only get the first row in the database, and there are 3 others that aren't displaying. Any assistance is much appreciated.
You need to use the following because if you call mysql_fetch_array outside of the loop, you're only returning an array of all the elements in the first row. By setting row to a new row returned by mysql_fetch_array each time the loop goes through, you will iterate through each row instead of whats actually inside the row.
while($row = mysql_fetch_array($result))
{
// This will loop through each row, now use your loop here
}
But the good way is to iterate through each row, as you have only three columns
while($row = mysql_fetch_assoc($result))
{
echo $row['name']." ";
echo $row['email']." ";
}
One common way to loop through results is something like this:
$result = mysql_query($query);
while ($row = mysql_fetch_assoc($result)) {
print_r($row);
// do stuff with $row
}
Check out the examples and comments on PHP.net. You can find everything you need to know there.
I have slow query problem, may be i am wrong, here is what i want,
i have to display more than 40 drop down lists at a single page with same fields , fetched by db, but i feel that the query takes much time to execute and also use more resources..
here is an example...
$sql_query = "SELECT * FROM tbl_name";
$rows = mysql_query($sql_query);
now i use while loop to print all records in that query in drop down list,
but i have to reprint same record in next drop down list up to 40 lists, so i use
mysql_data_seek() to move to first record and then reprint the next list and so on till 40 lists.
but this was seems slow to me so i use the second method like this same query for all 40 lists
$sql_query2 = "SELECT * FROM tbl_name";
$rows2 = mysql_query($sql_query2);
do you think that i wrong about the speed of query, or do you suggest me the another way that is faster than these methods....
Try putting the rows into an array like so:
<?php
$rows = array();
$fetch_rows = mysql_query("SELECT * FROM table");
while ($row = mysql_fetch_assoc($fetch_rows)) {
$rows[] = $row;
}
Then just use the $rows array in a foreach ($rows as $row) loop.
There is considerable processing overhead associated with fetching rows from a MySQL result resource. Typically it would be quite a bit faster to store the results as an array in PHP rather than to query and fetch the same rowset again from the RDBMS.
$rowset = array();
$result = mysql_query(...);
if ($result) {
while ($row = mysql_fetch_assoc($result)) {
// Append each fetched row onto $rowset
$rowset[] = $row;
}
}
If your query returns lots of rows (thousands or tens of thousands or millions) and you need to use all of them, you may reach memory limitations by storing all rows into an array in PHP. In that case it may be more memory-conservative to fetch rows individually from MySQL, but it will still probably be more CPU intensive.
Instead of printing the records, going back, and printing them again, put the records in one big string variable, then echo it for each dropdown.
$str = "";
while($row = mysql_fetch_assoc($rows)) {
// instead of echo...
$str .= [...];
}
// now for each dropdown
echo $str;
// will print all the rows.
I've been requesting MySQL results and looping through them like this:
$query = "SELECT * FROM $table";
$result = mysql_query($query);
for($i = 0; $i < mysql_num_rows($result); $i++){
echo mysql_result($result, $i, $row);
//do something else;
}
you can probably see what happens. What if a row has been deleted? What if the first item is gone? In that case, there could be 30 items in the list, but the last items index is at position 50. How do I fix this or what other systems can I use?
while($temp = mysql_fetch_assoc($result))
{
echo $temp['id']; // ID Column
}
But BTW. The mysql_num_rows returns the number of rows you fetched. If you have a row that was deleted, it wouldn't have been fetched, and therefore it wouldn't count towards the number.
Also - I believe the entire rowset is obtained at query time, rows deleted after the query is executed will still appear in your results.