The query SELECT * FROM TABLE WHERE id LIKE '%1% is not working properly, it's not select the id 1.
mysql_connect('localhost', 'root' , '');
mysql_select_db('database');
$sql = ("select * from search WHERE id LIKE '%3%'");
mysql_query($sql);
$my_variable = mysql_query($sql);
$display_data = mysql_fetch_row($my_variable);
while ($list = mysql_fetch_assoc($my_variable)) {
$id = $list['id'];
$title = $list['title'];
$keywords = $list['keywords'];
$img = $list['img'];
$link = $list['link'];
}
If you are looking to SELECT id 1 then use = not LIKE. The way LIKE is being used it will match every id that has a 1 in it and you are not guaranteed to get the first one in order, so instead use:
SELECT * FROM search WHERE id = 1
According to the PHP documentation of mysql_fetch_row it
Returns a numerical array that corresponds to the fetched row and moves the internal data pointer ahead.
Which means that the first result won't show up in the next (mysql_fetch_assoc) procedure. You could try removing the $display_data = mysql_fetch_row($my_variable); line and only use the while($list = mysql_fetch_assoc($my_variable)) { ... } procedure. See if that solves your problem.
$sql = ("select * from search WHERE id ='3'");
The id is an integer use = instead of like . Equal is more accurate.
And first echo your query in your program->
echo $sql;die;
copy that query and run it on your phpmyadmin
and then check is your column id is int type if it is then like will not give you the result. You have to use the where clause here .But if you have the column id is of type varchar then definitely give you the result .
Try to use search tab under your database->table in your phpmyadmin and put the condition there.
You will definitely get your answer there.
Related
after searching for several questions, i'm still struggling with mysql queries in PHP, my current goal is to do a MYSQL query that counts how many repeated strings are in a column and then return this amount in a INT variable to be written in the database.
The current code looks like:
//Fetch value from form and uppercase the string
$glitter = utf8_decode(strtoupper($_POST['code_string']));
$magic = mysql_query("SELECT COUNT(*) FROM table WHERE CODE_STR = '$glitter'");
The next step is inserting the var $magic into a INT field in the database, however the value is always 0.
Where is my mistake?
Thanks.
mysql_query returns a resource on success, or FALSE on error.
try this
$magic = mysql_query("SELECT COUNT(*) as count FROM table WHERE CODE_STR = '$glitter'");
$row = mysql_fetch_assoc($magic);
$count = $row['count'];
Your approach is correct. What you need to do now is
Change your query from
$magic = mysql_query("SELECT COUNT(*) FROM table WHERE CODE_STR = '$glitter'");
to
$magic = mysql_query("SELECT COUNT(*) as total_num FROM table WHERE CODE_STR = '$glitter'");
mysql_fetch_assoc() Use the returned value from table
$magic_row = mysql_fetch_assoc($magic);
echo $magic_row['total_num'];
See
http://php.net/manual/en/function.mysql-fetch-assoc.php
Ok, don't know if this is simple in practice as it is in theory but I want to know.
I have a single INSERT query were by in that query, i want to extract the AUTO_INCREMENT value then reuse it in the same query.
For example
//values to be inserted in database table
$a_name = $mysqli->real_escape_string($_POST['a_name']);
$details = $mysqli->real_escape_string($_POST['details']);
$display_type = $mysqli->real_escape_string($_POST['display_type']);
$getId = mysqli_insert_id();
//MySqli Insert Query
$insert_row = $mysqli->query("INSERT INTO articles (a_name,details,display_type,date_posted) VALUES('$a_name','$details','$display_type$getId',CURRENT_TIMESTAMP)");
Apparently, am getting a blank value(I know because the mysqli_insert_id() is before the query, but I've tried all i could but nothing has come out as i want. Can some please help me on how to achive this
From my knoweldge this cant be done. Because no query has been run, MySQL is unable to return the ID of said query.
You could use a classic approach, pull the id of the previous record and add 1 to it, this is not a great solution as if a record is deleted, the auto increment value and the last value +1 may differ.
Run multiple queries and then use the insert_id (MySQLi is different to what you are using, you are best using $db->lastInsertId(); as mentioned in the comments.
Run a query before hand and store it as a variable;
SELECT auto_increment FROM INFORMATION_SCHEMA.TABLES WHERE table_name = 'tablename'
I strongly recommend Option 2, it is simply the cleanest and most reliable method for what you are looking to achieve.
It seems the value required for $display_type is :$display_type + (max(id) + 1).
In order to get the max_id you'll have to do this query before :
$sql = "SELECT id FROM articles ORDER BY id DESC LIMIT 1";
$result = mysqli->query($sql);
$maxid = $result->fetch_array(MYSQLI_NUM);
// $maxid[0] will contains the value desired
// Remove the mysqli_insert_id() call - Swap $getid by ($maxid[0] + 1)
// and u're good to go
N.B. update the name of ur primary key in the query $sql.
EDIT :
Assuming the weakness of the query and the quick resarch i did.
Try to replace $sql by (don't forget to Update DatabaseName & TableName values) :
$sql = SELECT `AUTO_INCREMENT`
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_SCHEMA = 'DatabaseName'
AND TABLE_NAME = 'TableName';
That Should do it . More info on the link below :
Stackoverflow : get auto-inc value
I don't think this can be done. You'll have to first insert the row, then update display_type, in two separate queries.
Thanks guys for your opinions, out of final copy, paste, edit and fix; here is the final working code(solution)
`
//values to be inserted in database table
$a_name = $mysqli->real_escape_string($_POST['a_name']);
$details = $mysqli->real_escape_string($_POST['details']);
$display_type = $mysqli->real_escape_string($_POST['display_type']);
//Select AUTO_INCREMENT VALUE
$sql = "SELECT `AUTO_INCREMENT`
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_SCHEMA = 'chisel_bk'
AND TABLE_NAME = 'articles'";
$result = $mysqli->query($sql);
$maxid = $result->fetch_array(MYSQLI_NUM);
$getId = $maxid[0];
//MySqli Insert Query
$insert_row = $mysqli->query("INSERT INTO articles (a_name,details,display_type,date_posted) VALUES('$a_name','$details','$display_type$getId',CURRENT_TIMESTAMP)");
This happens to do the magic!!!
`
I want to get the maximum id of row data. In my table first column is id then firstname and etc.
this is the sql command I used to get the max(id) of row data.
<?PHP $maxid=$this->db->query("SELECT MAX(id) FROM `emplyee_personal_details`");
print_r( $maxid) ;?>
but it prints bunch of data as it is a array. But I need only the maximam id of row data for validation before data inserting and updating.
How to get the maxid. I use codeigniter framework.
Try this:
$maxid = $this->db->query('SELECT MAX(id) AS `maxid` FROM `emplyee_personal_details`')->row()->maxid;
UPDATE
This will work even if your table is empty (unlike my example above):
$maxid = 0;
$row = $this->db->query('SELECT MAX(id) AS `maxid` FROM `emplyee_personal_details`')->row();
if ($row) {
$maxid = $row->maxid;
}
The problem with using a raw query like "SELECT MAX(id) ..." is it is not abstract and may not work for every SQL engine. I think the active record way to do it is like this:
$this->db->select_max('id');
$query = $this->db->get('emplyee_personal_details');
// Produces: SELECT MAX(id) as age FROM emplyee_personal_details
See: http://ellislab.com/codeigniter/user-guide/database/active_record.html
SELECT id FROM table ORDER BY id DESC LIMIT 1
That will get you the highest id value, and when I read this, "it prints bunch of data as it is a array" I get the sense that what you really want is a part of that array. DB queries always return complex structures like arrays or objects. So if you wanted just the scalar value (the number as an integer) you might use something like this:
$maxid = (int)$maxid['id'];
or like this (if you have an object):
$maxid = (int)$maxid->id;
HTH, ~Ray
Try this,hope it helps,
return $this->db->select_max('id')
->get('your_table_name')
->row()->id;
public function getMaxCategoryId() {
$query = $this->db->query("SELECT category_id+1 AS maxid FROM " . DB_PREFIX . "category ORDER BY category_id DESC LIMIT 1");
return $query->row['maxid'];
}
error undefined index maxid
<?php`$qry = "select max(ID)+1 As ID from records";`
$result = $con->query($qry);
$row = $result->fetch_assoc();`echo "New ID To Enter = ".$row["ID"];?>
After Connection Just Write This Code It Will Work
I am having some difficulty running some SQL code.
What I am trying to do is, find a row that contains the correct username, and then get a value from that correct row.
This is my SQL in the php:
mysql_query("SELECT * FROM users WHERE joined='$username' GET name")
As you can see, it looks for a username in users and then once found, it must GET a value from the correct row.
How do I do that?
You need some additional PHP code (a call to mysql_fetch_array) to process the result resource returned by MySQL.
$result = mysql_query("SELECT name FROM users WHERE joined='$username'");
$row = mysql_fetch_array($result);
echo $row['name'];
mysql_query("SELECT `name` FROM users WHERE joined='$username' ")
Just select the right column in your 'select clause' like above.
Edit: If you are just starting out though, you might want to follow a tutorial like this one which should take you through a nice step by step (and more importantly up to date functions) that will get you started.
mysql_query("SELECT name FROM users WHERE joined='$username'")
$q = mysql_query("SELECT * FROM users WHERE joined='$username'");
$r = mysql_fetch_array($q);
$name = $r['user_name']; // replace user_name with the column name of your table
mysql_query("SELECT name FROM users WHERE joined='$username' ")
Read documentation : http://dev.mysql.com/doc/refman/5.0/en/select.html
I have this
$sql = "SELECT id FROM table";
$result = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_array($result)){
echo $row['id'];
}
This echo's all id's found in the table.
How can I choose to echo only a selected id.
Say the second id found on the table?
EDIT
I think I have confused people and myself aswell.
Let me try to explain again.
Using the above query I can echo all results found in the table with echo $row['id'];
However I do not want echo all results, just selected ones.
You guys have suggested I use limit or a Where clause.
If I do this I will be limited to just one record. This is not what I want.
I want to echo a selection of records.
Something likes this
echo $row['id'][5], $row['id'][6], $row['id'][6]
But obviously this is incorrect syntax and will not work but hopefully you get what I am trying to do.
Thanks
If you only want the second row then you could change your query to use offset and limit e.g.
SELECT id FROM table LIMIT 1, 1
You could also use a for loop instead of the while loop and then put in a conditional.
UPDATE
Just noticed comments above - you also need to sort the PHP bug by changing mysql_fetch_array to mysql_fetch_assoc.
UPDATE 2
Ok based on your update above you are looking to get all of the rows into an array which you can then iterate over.
You can just use mysql_fetch_array and then use $array[0]. For example:
$sql = "SELECT id FROM table";
$result = mysql_query($sql) or die(mysql_error());
$ids = array();
while($row = mysql_fetch_array($result)) {
$ids[] = $row[0];
}
From what I can gather from your questions you should not be selecting all records in the table if you wish to just use the Nth value, use:
SELECT id FROM table LIMIT N, 1
That will select the Nth value that was returned. Note: The first result is 0 so if you wish to get the second value the Nth value should be 1.
mysql_data_seek() let's you jump to a specific data-set(e.g. the 2.nd)
Example:
$sql = "SELECT id FROM table";
$result = mysql_query($sql) or die(mysql_error());
//get the 2nd id(counting starts at 0)
if(mysql_data_seek($result,1))
{
$row=mysql_fetch_assoc($result);
echo $row['id'];
}
OR:
use mysqli_result::fetch_all
It returns an array instead of a resultset, so you can handle it like an array and select single items directly (requires PHP5.3)