I'm trying to add a new mysql column in a table, using an insert_id from an insert of another table. This is the sentence that i use...
string sqlInsert = "INSERT INTO test (IdPico, Nombre, TextoBienvenida, FechaCreacion) VALUES (1, 'nombretest', 'aslkñdfa lsñdk asjd asldkf añlsj f', '2011-07-13 10:22:53'); ";
sqlInsert += "SET #IDTESTCREATED := CONCAT('Test', LAST_INSERT_ID(); ";
sqlInsert += "ALTER TABLE Usuarios ADD COLUMN #IDTESTCREATED BIT DEFAULT 0; ";
I using ASP.NET 4.0 and MySql connection, and server responds with 'Fatal error encountered during command execution. '
Could anybody help me?
Well ... I answer myself.
After making a deep search, I have not found how to add a column dynamically by a variable in mysql.
At end I had to make two querys, first to insert the test and get the id, and second to update the users table.
Since the insertion and retrieval of id are in the same query, no problems of persistent connections and concurrent updates.
string sqlInsert = "INSERT INTO Test (<fields>) VALUES (<values>);";
sqlInsert += "SELECT LAST_INSERT_ID() AS IdTestInserted; ";
string idnewtest = <result of insert query>;
string sqlAlter = "ALTER TABLE Users ADD COLUMN Test" + idnewtest + " BIT DEFAULT 0; ";
I regret not having found the answer, but at least I achieved my goal.
Thank you all for your help!
Related
I am trying to follow the following blog and use an UPDATE, IF, INSERT INTO statement seeing that it will no go over my data twice.
Please note it is for php.
The statement that I have is as follows
$query = "UPDATE
" . $this->table_name2 . "
SET
batch = :batch,
created = :created
WHERE
id = :id
IF row_count() = 0
INSERT INTO " . $this->table_name2 . "
SET
id=:id,
batch=:batch,
created=:created";
But my return value always comes back as false, and I do not know where the problem is.
If I try the first half of the statement it updates the information:
$query = "UPDATE
" . $this->table_name2 . "
SET
batch = :batch,
created = :created
WHERE
id = :id
And if I try the second half of the statement it INSERTS the information:
INSERT INTO " . $this->table_name2 . "
SET
id=:id,
batch=:batch,
created=:created";
I do not find any help after allot of searching so far, and I feel that somehow my IF statement may not be correct, but I do not even get info on the IF row_count() = 0 in the docs.
The IF statement is available in stored procedures only.
If id is the primary key (or unique) this should work for you:
INSERT INTO " . $this->table_name2 . "
SET
id = :id,
batch = :batch,
created = :created
ON DUPLICATE KEY UPDATE
batch = :batch,
created = :created
This will execute the INSERT statement only, of the ID does not exits yet. If it already exists, it will execute the UPDATE part.
I want my the id field in my table to be a bit more " random" then consecutive numbers.
Is there a way to insert something into the id field, like a +9, which will tell the db to take the current auto_increment value and add 9 to it?
Though this is generally used to solve replication issues, you can set an increment value for auto_increment:
auto_increment_increment
Since that is both a session and a global setting, you could simply set the session variable just prior to the insert.
Besides that, you can manually do it by getting the current value with MAX() then add any number you want and insert that value. MySQL will let you know if you try to insert a duplicate value.
You have a design flaw. Leave the auto increment alone and shuffle your query result (when you fetch your data)
As far as i know, it's not possible to 'shuffle' your current IDs. If you wanted though, you could pursue non-linear IDs in the future.
The following is written in PDO, there are mysqli equivalents.
This is just an arbitrary INSERT statement
$name = "Jack";
$conn = new PDO("mysql:host=$dbhost;dbname=$dbname",$dbuser,$dbpass);
$sql = "INSERT INTO tableName (name) VALUES(:name)";
$q = $conn->prepare($sql);
$q->execute(':name' => $name);
Next, we use lastInsertId() to return the ID of the last inserted row, then we concatenate the result to rand()
$lastID = $conn->lastInsertId();
$randomizer = $lastID.rand();
Finally, we use our 'shuffled' ID and UPDATE the previously inserted record.
$sql = "UPDATE tableName SET ID = :randomizer WHERE ID=:lastID ";
$q = $conn->prepare($sql);
$q->execute(array(':lastID' => $lastID , ':randomizer' => $randomizer));
An idea.. (Not tested)
CREATE TRIGGER 'updateMyAutoIncrement'
BEFORE INSERT
ON 'DatabaseName'.'TableName'
FOR EACH ROW
BEGIN
DECLARE aTmpValueHolder INT DEFAULT 0;
SELECT AUTO_INCREMENT INTO aTmpValueHolder
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_SCHEMA = 'DatabaseName'
AND TABLE_NAME = 'TableName';
SET NEW.idColumnName =aTmpValueHolder + 9;
END;
Edit : If the above trigger doesn't work try to update AUTO_INCREMENT value directly into the system's schema. But as noted by Eric, your design seems to be flawed. I don't see the point of having an auto-increment here.
Edit 2 : For a more 'random' and less linear number.
SET NEW.idColumnName =aTmpValueHolder + RAND(10);
Edit 3 : As pointed out by Jack Williams, Rand() produces a float value between 0 and 1.
So instead, to produce an integer, we need to use a floor function to transform the 'random' float into an integer.
SET NEW.idColumnName =aTmpValueHolder + FLOOR(a + RAND() * (b - a));
where a and b are the range of the random number.
I'm having a problem in retrieving last record id from database. This code below, is the closer I can get. But still, it return record id, as 0; ,then when I execute again, it will return, record of previous execute, not the current one.
sql = "insert into program (prog_det,budget,prog_obj,outcome,target_group,awareness,engagement,issue,seq_no) value ('"&prog_title&"','"&prog_budget&"','"&prog_obj&"','"&prog_result&"','"&prog_target&"','"&prog_aware&"','"&prog_involment&"','"&prog_issues&"','99');"
sql2 = "select last_insert_id() as last_id"
set kpi_prog_conn=Server.CreateObject("ADODB.Connection")
set kpi_prog_rs=Server.CreateObject("ADODB.Recordset")
kpi_prog_conn.Open ObjConn
kpi_prog_conn.Execute(sql)
kpi_prog_conn.Open sql2,objConn,adLockPessimistic
response.write kpi_prog_rs("last_id")
Your penultimate line looks wrong
Try
kpi_prog_rs.Open sql2,kpi_prog_conn,adLockPessimistic
IS the ID you are trying to retrieve is the Primary key of the corresponding table? Try using Scope_Identity() instead of last_insert_id()
Query - SELECT SCOPE_IDENTITY() AS [LAST_IDENTITY]
It returns you the last inserted id into the table
I don't have mysql but try:
sql2 = "select last_insert_id() as last_id;"
sql = "insert into program (prog_det,budget,prog_obj,outcome,target_group,awareness,engagement,issue,seq_no) value ('"&prog_title&"','"&prog_budget&"','"&prog_obj&"','"&prog_result&"','"&prog_target&"','"&prog_aware&"','"&prog_involment&"','"&prog_issues&"','99');" & sql2
set kpi_prog_conn=Server.CreateObject("ADODB.Connection")
kpi_prog_conn.Open ObjConn
set kpi_prog_rs = kpi_prog_conn.Execute(sql)
anotherRecordset = kpi_prog_rs.NextRecordset
response.write anotherRecordset("last_id")
I am trying to insert records in 2 different mysql tables. Here's the situation:
Table 1: is_main that contains records of resorts with a primary key called id.
Table 2: is_features that contains a list of features that a resort can have (i.e. beach, ski, spa etc...). Each feature has got a primary key called id.
Table 3: is_i2f to connect each resort id with the feature id. This table has got 2 fields: id_i and id_f. Both fields are primary key.
I have created a form to insert a new resort, but I'm stuck here. I need a proper mysql query to insert a new resort in the is_main table and insert in is_i2f one record for each feature it has, with the id of the resort id id_i and the id of the feature id id_f.
$features = ['beach','relax','city_break','theme_park','ski','spa','views','fine_dining','golf'];
mysql_query("INSERT INTO is_main (inv_name, armchair, holiday, sipp, resort, price, rooms, inv_length, more_info)
VALUES ('$name', '$armchair', '$holiday', '$sipp', '$resort', '$price', '$rooms', '$length', '$more_info')");
$id = mysql_insert_id();
foreach($features as $feature) {
if(isset($_POST[$feature])) {
$$feature = 1;
mysql_query("INSERT INTO is_i2f (id_i, id_f) VALUES (" . $id . ", ?????????????? /missing part here????/ ); }
else {
$$feature = 0; }
}
Thanks.
Please, I'm going CrAzY!!!!!!!!!!!!!!
This may not be relevant to you, but...
Would it not make more sense to leave the link table unpopulated? You can use JOINs to then select what you need to populate the various views etc in your application
i.e. query to get 1 resort with all features:
SELECT
Id,
f.Id,
f.Name
FROM IS_MAIN m
CROSS JOIN IS_FEATURES f
WHERE m.Id = $RequiredResortId
Please find the answer on Mysql insert into 2 tables.
If you want to do multiple insert at a time you can write a SP to fulfill your needs
If I understand you correctly you could concatenate variable amount of to be inserted/selected values into one query. (This is the second query which needs an id from the first.)
//initializing variables
$id = mysql_insert_id();
$qTail = '';
$i = -1;
//standard beginning
$qHead = "INSERT INTO `is_i2f` (`id`,`feature`) VALUES ";
//loop through variable amount of variables
foreach($features] as $key => $feature) {
$i++;
//id stays the same, $feature varies
$qValues[$i] = "('{$id}', '{$feature}')";
//multiple values into one string
$qTail .= $qValues[$i] . ',';
} //end of foreach
//concatenate working query, need to remove last comma from $qTail
$q = $qHead . rtrim($qTail, ',');
Now you should have a usable insert query $q. Just echo it and see how it looks and test if it works.
Hope this was the case. If not, sorry...
In my table I have an userID that is auto-incremented. In the same row I have an idHash. Is it possible to generate the idHash (simply an MD5 sum) from it directly with the same INSERT statement so that I don't have to SELECT the id, and then UPDATE the idHash again?
Problem is: I do not know the userID before it is being generated (auto-incremented) by MySQL.
Thanks
Frank
PS: I'm using PHP.
PPS: This question is all about a SINGLE INSERT. I know that I can use PHP or other languages to manually select the data and then update it.
I don't believe you can do it within a single INSERT statement.
What you probably could do is use an INSERT trigger, that both determines the new ID, hashes it, and then updates the record.
One solution I can recommend is using the last insert ID instead of re-querying the table. Here is a simplified example:
<?php
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$query = "INSERT INTO users VALUES (....)";
$mysqli->query($query);
$newUserID = $mysqli->insert_id;
$query = "UPDATE users SET idHash = MD5(userID) WHERE userID = $newUserID";
$mysqli->query($query);
/* close connection */
$mysqli->close();
?>
AFAIK there's no "secure" way for doing this in the same query if you're using auto_increment.
However, if rows are never deleted in your table, you can use this little trick :
insert into mytable (col1, col2, col3, idhash)
values ('', '', '', md5(select max(id) from mytable))
I don't understand why you need to hash the id though, why not use the id directly ?
This seems to work for me:
CREATE TABLE tbl (id INT PRIMARY KEY AUTO_INCREMENT, idHash TEXT);
INSERT INTO tbl (idHash) VALUES (MD5(LAST_INSERT_ID() + 1));
SELECT *, MD5(id) FROM tbl;
Note this will only work on single-row inserts as LAST_INSERT_ID returns the insert ID of the first row inserted.
Performing MD5(column_name) on an auto_increment value does not work as the value has not been generated yet, so it is essentially calling MD5(0).
PHP snippet
<?
$tablename = "tablename";
$next_increment = 0;
$qShowStatus = "SHOW TABLE STATUS LIKE '$tablename'";
$qShowStatusResult = mysql_query($qShowStatus) or die ( "Query failed: " . mysql_error() . "<br/>" . $qShowStatus );
$row = mysql_fetch_assoc($qShowStatusResult);
$next_increment = $row['Auto_increment'];
echo "next increment number: [$next_increment]";
?>
This will get you the next auto-increment and then you can use this in your insert.
Note: This is not perfect (Your method is imperfect as you will effectively have 2 primary keys)
From: http://blog.jamiedoris.com/geek/560/