[英]How to get table name from mysql result?
MySQL查詢如下
SELECT t1.*, t2.*, t3.*, t4.*, t5.*, t6.* FROM table1 t1 INNER JOIN table2 t2 INNER JOIN table3 t3 INNER JOIN table4 t4 INNER JOIN table5 t5 INNER JOIN table6 t6 order by t1.updated_time, t2.updated_time, t3.updated_time, t4.updated_time, t5.updated_time, t6.updated_time desc
從上面的查詢中,我需要帶有各自表名稱的結果,例如
Array( [0] => stdClass Object ( [id] => 1 [cloumn1] => data1 [column2] => table3 [updated_time] => data1 ) [1] => stdClass Object ( [id] => 2 [cloumn1] => data1 [column2] => table1 [updated_time] => data2 ) )
該表有超過15列,可能會有所不同。
如何通過修改查詢來達到結果?
查詢后,您可以進行foreach循環以獲取所有列名。 像這樣
while($row = mysqli_fetch_assoc($query)){
foreach($row as $key => $value){
echo "$key=$value";
}
}
Update
如果要在數據庫中獲取表名,可以嘗試類似的方法。
$sql = "SHOW TABLES FROM database";
$result = mysqli_query($conn,$sql);
while ($row = mysqli_fetch_row($result)) {
echo "Table: {$row[0]}\n";
}
更新
首先,我們需要獲取表名。 由於您不知道表名,因此可以使用以下代碼獲取表名
$tables = array();
$sql = "SHOW TABLES FROM database";
$result = mysqli_query($conn,$sql);
while ($row = mysqli_fetch_row($result)) {
$tables[] = $row[0];
}
如果有表,可以跳過上面的代碼將它們添加到數組中
$tables = array("table1","table2","table3");
獲取所有表后,我們可以從數據庫開始
$data = array();
foreach($tables as $table){
$query = "select * from $table";
$res = mysqli_query($conn,$query);
while($row = mysqli_fetch_assoc($res)){
$i=1;
foreach($row as $key => $value){
$data[$i][$key][$value];
$i++;
}
}
}
根據您的需要更新代碼。 這是一個例子
另一個解決方案(由某些ORM使用)
<?php $tables = ['table1', 'table2', 'table3']; ?>
http://dev.mysql.com/doc/refman/5.7/en/show-columns.html
<?php
$structures = [];
$structuresLinear = [];
foreach( $tables as $table ) {
$query = mysqli_query('SHOW TABLES FROM ' . $table);
while ($row = mysqli_fetch_row($query)) {
$structures[$table][] = $row[0];
$structuresLinear[] = sprintf('%s.%s as %s', $table, $row[0], $table . '_' . $row[0]);
}
}
您將在$structures
為每個表保留所有字段,並在另一個$structuresLinear
上重命名不同的字段。
例如: ['table1.id as table1_id', 'table1.name as table1_name', ...]
<?php
$sql = 'SELECT ';
$sql .= implode(', ', $structuresLinear);
$sql .= ' FROM ' . implode(' INNER JOIN ', $tables);
$sql .= ' ORDER BY ' . implode(', ',
array_map(function($t) {
return $t . '.updated_time';
}, $tables)
);
?>
您將擁有以下內容:
SELECT
table1.id as table1_id,
table2.id as table2_id, table2.col2 as table2_col2, table2.col3 as table2_col3
FROM table1
INNER JOIN table2
ORDER BY
table1.updated_time,
table2.updated_time
最后的數組將是:
Array(
[0] => Array(
table1_id = table1.id,
table2_id = table2.id,
table2_col2 = table2.col2,
table2_col3 = table2.col3,
)
)
您還可以拆分每個$ key結果以包含表或創建multidim數組:)
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