mysql从另一个表中更新列的值

我有两张桌子，两个都像

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

我需要根据每个表中的检查名称将tableA的值的value复制到tableB 。

这个UPDATE语句的任何提示？

Answer 1

除了这个答案，如果您需要根据t2.valuedynamic更改t1.value，您可以执行例如：

 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' 

Answer 2

你需要join这两个表格：

例如你想把tableA中的name的值复制到tableB ，并且他们有相同的ID

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

更新1

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

更新2

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

Answer 3

第二种可能性是，

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

Answer 4

第二个选项也是可行的，如果你正在使用安全更新模式（并且你得到一个错误，指示你已经尝试更新一个没有使用KEY列的WHERE表），通过添加：

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

Answer 5

 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(); } }