I'am trying to create table with variable name and columns via forms here's my query
$execute = mysql_query('CREATE TABLE '.$tName.'(
'.$cName.' '.$cType.' )');
And it's creating table with name from variable $tName but it doesnt create the columns.
Try this:
$sql= "CREATE TABLE $tName($cName $cType(45),$cName $cType(50))";
//echo $sql;
$op =mysql_query($sql);
OK try this, I think it gives the general idea
<?php
include 'connect-db.php';
$tbl_name='kerry';
$field1='name';
$field2='value';
$field3='date';
mysql_query("CREATE TABLE `{$tbl_name}`
(
$field1 varchar(15),
$field2 int,
$field3 date
)");
?>
Related
I have got MySQL table with three columns 'primary Key','debit_cash','user_id' So now i want to update the debit_cash values to the corresponding user_id by adding "15" to the value already present. The debit_cash is in VARCHAR so i tried converting to int and sum it , but still the values in MySQL is not changing .
Here is my code:
<?php
if($_SERVER['REQUEST_METHOD']=='POST'){
//Getting values
$user_id = $_POST['user_id'];
//importing database connection script
require_once('dbConnect.php');
//Creating sql query
$sql = "SELECT cos_details.debit_cash AS debitCash,
(convert(int, debit_cash)+15) AS updatedDebitCash
FROM cos_details
UPDATE cos_details SET debit_cash = '$updatedDebitCash'
WHERE user_id = $user_id";
//Updating database table
if(mysqli_query($con,$sql)){
echo 'Updated Successfully';
}else{
echo 'Could Not Update Try Again';
}
//closing connection
mysqli_close($con);
}
Any one please help me.
Seems that you don't need the select but the update only
UPDATE cos_details SET debit_cash = cast( (convert(int, debit_cash)+15) as VARCHAR(20))
WHERE user_id = $user_id
could be that your user_id is a string too so you should surround the value with quote
UPDATE cos_details SET debit_cash = cast( (convert(int, debit_cash)+15) as VARCHAR(20))
WHERE user_id = '$user_id'
attent table :
id int(11) primary key
userID int (11) forigen key from users table
date date
start_time text
end_time text
approv enum default 0
my query :
$sql = "INSERT INTO attent ".
"(id,userID,date,start_time,end_time,approv) ".
"VALUES ".
"('NULL','$userid','$date','$start_time','$end_time','NULL')";
$query = mysqli_query($db,$sql);
I got config.php file that help communicate with dB. I did a few insert queries before and all of them work just fine. I can not understand where is my mistake.
Write your query as below:-
"INSERT INTO attent(`id`,`userID`,`date`,`start_time`,`end_time`,`approv`)
VALUES(NULL,'$userid','$date','$start_time','$end_time',NULL)"
Try like this use back tick,because date is reserved word in mysql
$sql = "INSERT INTO `attent` ".
"(`id`,`userID`,`date`,`start_time`,`end_time`,`approv`) ".
"VALUES ".
"('','$userid','$date','$start_time','$end_time','')";
It's easier to understand what's going on if you will try to put such a query manually through e.g. phpMyAdmin
I don't know your table structure but try to skip id attribute as it's often autoincrementable. You can always skip the fields that is not set to NOT NULL.
Like that:
$sql = "INSERT INTO `attent` ".
"(`userID`,`date`,`start_time`,`end_time`,`approv`) ".
"VALUES ".
"('$userid','$date','$start_time','$end_time','')";
For the life of me I cannot figure out why my update statement will not update the table row but instead it creates a new row. I have an ID column that is the unique identifier and is auto_increment, I am just not sure if you can update an auto_incremented data set the way i am trying to.
I have a form that is echo'ing data from the database into the fields and then am using it to edit the fields and update them.
The code:
<?php
$EntryID = $_GET['Eid'];
$IDlist = mysql_query("SELECT * FROM BD WHERE Id='$EntryID'");
$IDresults = mysql_fetch_array($IDlist);
$update_query = "UPDATE `BD` SET `Id` ='$IDresults['Id']',`EntryTitle` = '$MyTitle',`EntryDescription` = '$MyDescription',`Category` = '$MyCategory' WHERE `Id` ='$EntryID'";
mysql_query($update_query);
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
else{
header('location: /admin/bd-edit-entry.php?sub=1');
exit();
}
mysql_close($con);
?>
Any help or advice would be a great.
SET `Id` ='$IDresults['Id']'
should be either:
SET `Id` ='$IDresults[Id]'
or
SET `Id` ='{$IDresults['Id']}'
If you turn on error reporting, you should get errors about a bad index.
Or you can leave this column out of the update entirely, since this column isn't changing.
I have a table with id which is the primary key and user_id which is a foreign key but the session is based on this in my code.
I have tried EVERYTHING, so I will post my full code.
The form should insert if there is not a user_id with the same session_id in the table. If there is, it should update.
At the moment, when the user has not visited the form before (no user_id in the table) and data is inserted in, the page returns to the location page: but the data is not inserted in the table. if the user changes the data once it is updated it doesn't change either.
This is the table structure:
`thesis` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`user_id` int(11) NOT NULL,
`thesis_Name` varchar(200) NOT NULL,
`abstract` varchar(200) NOT NULL,
`complete` int(2) NOT NULL DEFAULT '1',
PRIMARY KEY (`id`),
KEY `user_id` (`user_id`)
)
The code I have been using (and failing):
$err = array();
$user_id = intval($_SESSION['user_id']);
// otherwise
if (isset($_POST['doThesis'])) {
$link = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die("Couldn't make connection.");
// check if current user is banned
$the_query = sprintf("SELECT COUNT(*) FROM users WHERE `banned` = '0' AND `id` = '%d'",
$user_id);
$result = mysql_query($the_query, $link);
$user_check = mysql_num_rows($result);
// user is ok
if ($user_check > 0) {
// required field name goes here...
$required_fields = array('thesis_Name','abstract');
// check for empty fields
foreach ($required_fields as $field_name) {
$value = trim($_POST[$field_name]);
if (empty($value)) {
$err[] = "ERROR - The $field_name is a required field" ;
}
} // no errors
if (empty($err)) {
$id = mysql_real_escape_string($_POST['id']);
$thesis_Name = mysql_real_escape_string($_POST['thesis_Name']);
$abstract = mysql_real_escape_string($_POST['abstract']);
//replace query
$query = "REPLACE INTO thesis ( thesis_Name, abstract) VALUES ('$thesis_Name',
'$abstract') where id='$_SESSION[user_id]'";
if (!mysql_query($the_query))
echo "the query failed";
else header ("location:myaccount.php?id=' . $user_id");
}}}
$rs_settings = mysql_query("SELECT * from thesis WHERE user_id = $user_id;");
?>
<br>
<form action="thesis.php" method="post" name="regForm" id="regForm" >
class="forms">
<?php
$num_rows = mysql_num_rows($rs_settings);
if($num_rows > 0) { ?>
<?php while ($row_settings = mysql_fetch_array($rs_settings)) {?>
Title of Proposed Thesis<span class="required">*</span>
<textarea name="thesis_Name" type="text" style="width:500px; height:150px"
id="thesis_Name" size="600"><?php echo $row_settings['thesis_Name']; ?> </textarea>
</tr>
<tr>
<td>Abstract<span class="required">*</span>
</td>
<td><textarea name="abstract" style="width:500px; height:150px"
type="text" id="abstract" size="600"><?php echo $row_settings['abstract']; ?>
</textarea></td>
</tr>
<?php }
} else { ?>
//shows fields again without echo
I've tried var_dum($query) but nothing appears
PS I know the code isn't perfect but I'm not asking about this right now
I can't see how your replace statement will ever insert the initial row, as the where clause is always going to be false (there won't be a row with that user Id).
I think of you want to use replace you need to replace into thesis (id, userid, etc) without a where clause. If id and userid have a unique constraint and a row for userid exists then it will be updated; if it doesn't exist it will be inserted.
However- if you don't know id- which you won't if you are using auto increment, then I'm not sure you can do this with replace. See http://dev.mysql.com/doc/refman/5.0/en/replace.html
Why don't you check for the existence of a row an then use update or insert?
BTW, is the idea that a user can enter multiple theses into a form, or just one? Your table suggests they can have multiple. If this is what you are trying to achieve then I think you should be storing the id of each thesis in a hidden field as part of the form data. You would then be able to use REPLACE INTO thesis (id, user_id, thesis_name, abstract) VALUES ($id, $user_id, $thesis_name, $abstract) where id is the id of the thesis obtained from each hidden field. If this is not present, i.e. the user has entered a new thesis, then use NULL for id in the insert. This will work using the REPLACE INTO as the id column is auto increment.
Perhaps you mean user_id not id:
$query = "REPLACE INTO thesis ( thesis_Name, abstract)
VALUES ('$thesis_Name','$abstract')
WHERE user_id='{$_SESSION['user_id']}'";
Or if you do mean the id from $_POST['id']
$query = "REPLACE INTO thesis ( thesis_Name, abstract)
VALUES ('$thesis_Name','$abstract')
WHERE id='$id'";
Also instead of REPLACE you should use UPDATE. Im pretty sure its faster because REPLACE basically deletes the row then inserts it again, im pretty sure you need all the fields and values else your insert default values. From the manual:
Values for all columns are taken from the values specified in the
REPLACE statement. Any missing columns are set to their default
values, just as happens for INSERT
So you should use:
$query = "UPDATE thesis
SET thesis_Name='$thesis_Name', abstract='$abstract'
WHERE id='$id'";
You are doing everything right just one thing you are doing wrong
Your replace query variable is $query and you executing $the_query.
you wrong here:
$query = "REPLACE INTO thesis ( thesis_Name, abstract) VALUES ('$thesis_Name',
'$abstract') where id='$_SESSION[user_id]'";
if (!mysql_query($the_query)) // this is wrong
echo "the query failed";
replace it with:
$query = "REPLACE INTO thesis ( thesis_Name, abstract) VALUES ('$thesis_Name',
'$abstract') where id='$_SESSION[user_id]'";
if (!mysql_query($query)) // use $query
echo "the query failed";
Suppose I have a table called "device" as below:
device_id(field)
123asf15fas
456g4fd45ww
7861fassd45
I would like to use the code below to insert new record:
...
$q = "INSERT INTO $database.$table `device_id` VALUES $device_id";
$result = mysql_query($q);
...
I don't want to insert a record that is already exist in the DB table, so how can I check whether it have duplicated record before inserting new record?
Should I revise the MYSQL statement or PHP code?
Thanks
UPDATE
<?php
// YOUR MYSQL DATABASE CONNECTION
$hostname = 'localhost';
$username = 'root';
$password = '';
$database = 'device';
$table = 'device_id';
$db_link = mysql_connect($hostname, $username, $password);
mysql_select_db( $database ) or die('ConnectToMySQL: Could not select database: ' . $database );
//$result = ini_set ( 'mysql.connect_timeout' , '60' );
$device_id = $_GET["device_id"];
$q = "REPLACE INTO $database.$table (`device_id`) VALUES ($device_id)";
$result = mysql_query($q);
if (!$result) {
die('Invalid query: ' . mysql_error());
}
?>
Since I understood well your question you have two ways to go, it depends how you would like to do the task.
First way -> A simple query can returns a boolean result in the device_id (Exists or not) from your database table. If yes then do not INSERT or REPLACE (if you wish).
Second Way -> You can edit the structure of your table and certify that the field device_id is a UNIQUE field.
[EDITED]
Explaining the First Way
Query your table as follow:
SELECT * FROM `your_table` WHERE `device_id`='123asf15fas'
then if you got results, then you have already that data stored in your table, then the results is 1 otherwise it is 0
In raw php it looks like:
$result = mysql_query("SELECT * FROM `your_table` WHERE `device_id`='123asf15fas'");
if (!$result)
{
// your code INSERT
$result = mysql_query("INSERT INTO $database.$table `device_id` VALUES $device_id");
}
Explaining the Second Way
If your table is not yet populated you can create an index for your table, for example go to your SQL command line or DBMS and do the follow command to your table:
ALTER TABLE `your_table` ADD UNIQUE (`device_id`)
Warning: If it is already populated and there are some equal data on that field, then the index will not be created.
With the index, when someone try to insert the same ID, will get with an error message, something like this:
#1062 - Duplicate entry '1' for key 'PRIMARY'
The best practice is to use as few SQL queries as possible. You can try:
REPLACE INTO $database.$table SET device_id = $device_id;
Source