sql – mysql update column with value from another table

The Question :

240 people think this question is useful

I have two tables, both looking like

id  name  value
===================
1   Joe     22
2   Derk    30

I need to copy the value of value from tableA to tableB based on check name in each table.

Any tips for this UPDATE statement?

The Question Comments :

The Answer 1

451 people think this answer is useful

In addition to this answer if you need to change tableB.value according to tableA.value dynamically you can do for example:

UPDATE tableB
INNER JOIN tableA ON tableB.name = tableA.name
SET tableB.value = IF(tableA.value > 0, tableA.value, tableB.value)
WHERE tableA.name = 'Joe'

The Answer 2

168 people think this answer is useful

you need to join the two tables:

for instance you want to copy the value of name from tableA into tableB where they have the same ID

UPDATE tableB t1 
        INNER JOIN tableA t2 
             ON t1.id = t2.id
SET t1.name = t2.name 
WHERE t2.name = 'Joe'

UPDATE 1

UPDATE tableB t1 
        INNER JOIN tableA t2 
             ON t1.id = t2.id
SET t1.name = t2.name 

UPDATE 2

UPDATE tableB t1 
        INNER JOIN tableA t2 
             ON t1.name = t2.name
SET t1.value = t2.value

The Answer 3

99 people think this answer is useful

Second possibility is,

UPDATE TableB 
SET TableB.value = (
    SELECT TableA.value 
    FROM TableA
    WHERE TableA.name = TableB.name
);

The Answer 4

4 people think this answer is useful
    UPDATE    cities c,
          city_langs cl
    SET       c.fakename = cl.name
   WHERE     c.id = cl.city_id

The Answer 5

3 people think this answer is useful

The second option is feasible also if you’re using safe updates mode (and you’re getting an error indicating that you’ve tried to update a table without a WHERE that uses a KEY column), by adding:

UPDATE TableB  
SET TableB.value = (  
SELECT TableA.value  
    FROM TableA  
    WHERE TableA.name = TableB.name  
)  
**where TableB.id < X**  
;

The Answer 6

1 people think this answer is useful

Store your data in temp table

Select * into tempTable from table1

Now update the column

 UPDATE table1
    SET table1.FileName = (select FileName from tempTable where tempTable.id = table1.ID);

The Answer 7

0 people think this answer is useful

In my case, the accepted solution was just too slow. For a table with 180K rows the rate of updates was about 10 rows per second. This is with the indexes on the join elements.

I finally resolved my issue using a procedure:

CREATE DEFINER=`my_procedure`@`%` PROCEDURE `rescue`()
BEGIN
    declare str VARCHAR(255) default '';
    DECLARE n INT DEFAULT 0;
    DECLARE i INT DEFAULT 0;
    DECLARE cur_name VARCHAR(45) DEFAULT '';
    DECLARE cur_value VARCHAR(10000) DEFAULT '';
    SELECT COUNT(*) FROM tableA INTO n;
    SET i=0;
    WHILE i<n DO 
      SELECT namea,valuea FROM tableA limit i,1 INTO cur_name,cur_value;
      UPDATE tableB SET nameb=cur_name where valueb=cur_value;
      SET i = i + 1;
    END WHILE;

END

I hope it will help someone in the future like it helped me

The Answer 8

-5 people think this answer is useful

If you have common field in both table then it’s so easy !….

Table-1 = table where you want to update. Table-2 = table where you from take data.

  1. make query in Table-1 and find common field value.
  2. make a loop and find all data from Table-2 according to table 1 value.
  3. again make update query in table 1.

$qry_asseet_list = mysql_query("SELECT 'primary key field' FROM `table-1`");

$resultArray = array();
while ($row = mysql_fetch_array($qry_asseet_list)) {
$resultArray[] = $row;
}



foreach($resultArray as $rec) {

    $a = $rec['primary key field'];

    $cuttable_qry = mysql_query("SELECT * FROM `Table-2` WHERE `key field name` = $a");

    $cuttable = mysql_fetch_assoc($cuttable_qry);



    echo $x= $cuttable['Table-2 field']; echo " ! ";
    echo $y= $cuttable['Table-2 field'];echo " ! ";
    echo $z= $cuttable['Table-2 field'];echo " ! ";


    $k = mysql_query("UPDATE `Table-1` SET `summary_style` = '$x', `summary_color` = '$y', `summary_customer` = '$z' WHERE `summary_laysheet_number` = $a;");

    if ($k) {
        echo "done";
    } else {
        echo mysql_error();
    }


}

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