When I echo a variable it gave me a great looking list but when I update mysql database it only add the last row in the list I read that there is a problem with array and foreach with mysql.
foreach ($array["product"] as $list)
{
$ok = ''.$list[0].'';
$sql = mysql_query("UPDATE product SET desc='$ok' WHERE id='$id'") or die(mysql_error());
}
}
Is there an easy way to put my list in mysql whitout having to serialize I read its not the best thing to do and I am not familiar with that.
Thanks!
Related
I want retrieve data from mysql database and store in an array. Later I would display the data in html.
Below is my code snippet:
select co_no from faculty_co where course_code='$course1_code';
The output will display a total 5 co_no values. I want to store this values in an array and display as dropdown menu in html using select tag.
I am new to php. Please tell me how do I retrieve and store it.
I'm 'old school', and not much of a programmer, so I tend to do things procedurally.
I do it something like this:
include('path/to/connection/stateme.nts'); // the connection is a variable '$db'
$query = "my query";
$result = mysqli_query($db,$query);
$rows = array(); //initialise array
while($row = mysqli_fetch_assoc($result)){
$rows[] = $row;
}
Now you can store that as a json_encoded string or just retain it as an array for later use.
First off, php.net has a lot of documentation on mysql and other things and is pretty easy to understand.
Check out the documentation and examples here.
I'm trying to create a iterate through a row in a MySQL table in one of my CodeIgniter projects, how could I "loop" through the table? Is it just a simple for loop like in other languages?
EDIT:
The answer is as followed:
$query = $this->db->get('mytable'); // select table "mytable" from database
foreach ($query->result() as $row) { // loop thru table and access each row's field
// by using $row->fieldname
}
Maybe something like this
//$this->db->limit(10); // Optional if you want to limit, read about it
$result = $this->db->get('server'); //return all rows
foreach ($result as $row) {
$row->status = 'inactive'; // change value of status attribute or whatever
$this->db->update('server', $row)
}
Or maybe use $this->db->update_batch(); to update a stack of rows at one time.
I encourage you to read the CI database class documentation too.
Another suggestion, is to do all the business logic inside a model instead of a controller. But it's a matter of personal preference maybe.
I'm having a little problem with the codes given below. When I'm using the name="staff_number[]" then it insert the record with everything ok even if it is already in the database table and when i use name="staff_number" it does check the record and also give me alert box but when insert the record if it is not in the database it stores only the first number of the staff number like the staff no is 12345 it stores only 1. can anyone help in this record i think there is only a minor issue what I'm not able to sort out.
PHP Code:
<select placeholder='Select' style="width:912px;" name="staff_number[]" multiple />
<?php
$query="SELECT * FROM staff";
$resulti=mysql_query($query);
while ($row=mysql_fetch_array($result)) { ?>
<option value="<?php echo $row['staff_no']?>"><?php echo $row['staff_name']?></option>
<?php } ?>
</select>
Mysql Code:
$prtCheck = $_POST['staff_number'];
$resultsa = mysql_query("SELECT * FROM staff where staff_no ='$prtCheck' ");
$num_rows = mysql_num_rows($resultsa);
if ($num_rows > 0) {
echo "<script>alert('Staff No $prtCheck Has Already Been Declared As CDP');</script>";
$msg=urlencode("Selected Staff ".$_POST['st_nona']." Already Been Declared As CDP");
echo'<script>location.href = "cdp_staff.php?msg='.$msg.'";</script>';
}
Insert Query
$st_nonas = $_POST['st_nona'];
$t_result = $_POST['st_date'];
$p_result = $_POST['remarks'];
$arrayResult = explode(',', $t_result[0]);
$prrayResult = explode(',', $p_result[0]); $arrayStnona = $st_nonas;
$countStnona = count($arrayStnona);
for ($i = 0; $i < $countStnona; $i++) {
$_stnona = $arrayStnona[$i];
$_result = $arrayResult[$i];
$_presult = $prrayResult[$i];
mysql_query("INSERT INTO staff(st_no,date,remarks)
VALUES ('".$_stnona."', '".$_result."', '".$_presult."')");
$msg=urlencode("CDP Staff Has Been Added Successfully");
echo'<script>location.href = "cdp_staff.php?msg='.$msg.'";</script>';
}
Your $_POST['staff_number'] is actually an array.
So you have to access it like $_POST['staff_number'][0] here, 0 is a index number.
If the name of select is staff_number[] then $prtCheck will be a array so your check query must be in a loop to make sure your check condition.
if the name is staff_number then the below code is fine.
The answer of amit is right but I will complete it.
Your HTML form give to your PHP an array due to the use of staff_number[] with [] that it seems legit with the "multiple" attribute.
So you have to loop on the given values, you do it with a for and a lot of useless variables without really checking it. From a long time, we have the FOREACH loop structure.
I could help you more if i know what is the 'st_nona', st_date' and 'remarks' values.
According to your question you are getting difficulty in storing the data. This question is related to $_POST array.
Like your question we have selected following ids from the select : 1,2,3,4
It is only storing 1.
This is due to you have not used the loop when inserting the data.
Like below:
<?php
foreach($_POST['staffnumber'] as $staffnumber){
$query=mysql_query("select * from staff where staff_number =".$staffnumber);
if(mysql_num_rows($query)>0){
//action you want to perform
}else{
//action you want to perform like entering records etc. as your wish
}
}
?>
And I would like to suggest you that use the unique keys in database for field and use PHP PDO for database, as it is secure and best for OOPs.
Let me know if you have any queries.
I have one problem here, and I don't even have clue what to Google and how to solve this.
I am making PHP application to export and import data from one MySQL table into another. And I have problem with these tables.
In source table it looks like this:
And my destination table has ID, and pr0, pr1, pr2 as rows. So it looks like this:
Now the problem is the following: If I just copy ( insert every value of 1st table as new row in second) It will have like 20.000 rows, instead of 1000 for example.
Even if I copy every record as new row in second database, is there any way I can fuse rows ? Basically I need to check if value exists in last row with that ID_, if it exist in that row and column (pr2 for example) then insert new row with it, but if last row with same ID_ does not have value in pr2 column, just update that row with value in pr2 column.
I need idea how to do it in PHP or MySQL.
So you got a few Problems:
1) copy the table from SQL to PHP, pay attention to memory usage, run your script with the PHP command Memory_usage(). it will show you that importing SQL Data can be expensive. Look this up. another thing is that PHP DOESNT realese memory on setting new values to array. it will be usefull later on.
2)i didnt understand if the values are unique at the source or should be unique at the destination table.. So i will assume that all the source need to be on the destination as is.
I will also assume that pr = pr0 and quant=pr1.
3) you have missmatch names.. that can also be an issue. would take care of that..also.
4) will use My_sql, as the SQL connector..and $db is connected..
SCRIPT:
<?PHP
$select_sql = "SELECT * FROM Table_source";
$data_source = array();
while($array_data= mysql_fetch_array($select_sql)) {
$data_source[] = $array_data;
$insert_data=array();
}
$bulk =2000;
foreach($data_source as $data){
if(isset($start_query) == false)
{
$start_query = 'REPLACE INTO DEST_TABLE ('ID_','pr0','pr1','pr2')';
}
$insert_data[]=implode(',',$data).',0)';// will set 0 to the
if(count($insert_data) >=$bulk){
$values = implode('),(',$insert_data);
$values = substr(1,2,$values);
$values = ' VALUES '.$values;
$insert_query = $start_query.' '.$values;
$mysqli->query($insert_query);
$insert_data = array();
} //CHECK THE SYNTAX IM NOT SURE OF ALL OF IT MOSTLY THE SQL PART>> SEE THAT THE QUERY IS OK
}
if(count($insert_data) >=$bulk) // IF THERE ARE ANY EXTRA PIECES..
{
$values = implode('),(',$insert_data);
$values = substr(1,2,$values);
$values = ' VALUES '.$values;
$insert_query = $start_query.' '.$values;
$mysqli->query($insert_query);
$insert_data = null;
}
?>
ITs off the top off my head but check this idea and tell me if this work, the bugs night be in small things i forgot with the QUERY structure, print this and PASTE to PHPmyADMIN or you DB query and see its all good, but this concept will sqve a lot of problems..
$url = mysql_real_escape_string($_POST['url']);
$shoutcast_url = mysql_real_escape_string($_POST['shoutcast_url']);
$site_name = mysql_real_escape_string($_POST['site_name']);
$site_subtitle = mysql_real_escape_string($_POST['site_subtitle']);
$email_suffix = mysql_real_escape_string($_POST['email_suffix']);
$logo_name = mysql_real_escape_string($_POST['logo_name']);
$twitter_username = mysql_real_escape_string($_POST['twitter_username']);
with all those options in a form, they are pre-filled in (by the database), however users can choose to change them, which updates the original database. Would it be better for me to update all the columns despite the chance that some of the rows have not been updated, or just do an if ($original_db_entry = $possible_new_entry) on each (which would be a query in itself)?
Thanks
I'd say it doesn't really matter either way - the size of the query you send to the server is hardly relevant here, and there is no "last updated" information for columns that would be updated unjustly, so...
By the way, what I like to do when working with such loads of data is create a temporary array.
$fields = array("url", "shoutcast_url", "site_name", "site_subtitle" , ....);
foreach ($fields as $field)
$$field = mysql_real_escape_string($_POST[$field]);
the only thing to be aware of here is that you have to be careful not to put variable names into $fields that would overwrite existing variables.
Update: Col. Shrapnel makes the correct and valid point that using variable variables is not a good practice. While I think it is perfectly acceptable to use variable variables within the scope of a function, it is indeed better not use them at all. The better way to sanitize all incoming fields and have them in a usable form would be:
$sanitized_data = array();
$fields = array("url", "shoutcast_url", "site_name", "site_subtitle" , ....);
foreach ($fields as $field)
$sanizited_data[$field] = mysql_real_escape_string($_POST[$field]);
this will leave you with an array you can work with:
$sanitized_data["url"] = ....
$sanitized_data["shoutcast_url"] = ....
Just run a single query that updates all columns:
UPDATE table SET col1='a', col2='b', col3='c' WHERE id = '5'
I would recommend that you execute the UPDATE with all column values. It'd be less costly than trying to confirm that the value is different than what's currently in the database. And that confirmation would be irrelevant anyway, because the values in the database could change instantly after you check them if someone else updates them.
If you issue an UPDATE against MySQL and the values are identical to values already in the database, the UPDATE will be a no-op. That is, MySQL reports zero rows affected.
MySQL knows not to do unnecessary work during an UPDATE.
If only one column changes, MySQL does need to do work. It only changes the columns that are different, but it still creates a new row version (assuming you're using InnoDB).
And of course there's some small amount of work necessary to actually send the UPDATE statement to the MySQL server so it can compare against the existing row. But typically this takes only hundredths of a millisecond on a modern server.
Yes, it's ok to update every field.
A simple function to produce SET statement:
function dbSet($fields) {
$set='';
foreach ($fields as $field) {
if (isset($_POST[$field])) {
$set.="`$field`='".mysql_real_escape_string($_POST[$field])."', ";
}
}
return substr($set, 0, -2);
}
and usage:
$fields = explode(" ","name surname lastname address zip fax phone");
$query = "UPDATE $table SET ".dbSet($fields)." WHERE id=$id";