Question

我可以从命令行输出文本文件中的SHOW ENGINE INNODB STATUS\G;。还有使用mysqli_fetch_array（）或mysqli_fetch_assoc()

的任何查询

以下也是可能的，因为我们有列名（类型，名称，状态）

mysql> SHOW ENGINE PERFORMANCE_SCHEMA STATUS;
+--------------------+-------------------------------------------------------------+----------+
| Type               | Name                                                        | Status   |
+--------------------+-------------------------------------------------------------+----------+
| performance_schema | events_waits_current.size                                   | 186      |

现在，我想通过php在Web浏览器中打印SHOW ENGINE INNODB STATUS\G;的完整输出。

我已经通过了var_dump()和serialize()，但却无法真正开始工作。请帮忙

网络浏览器中的预期输出

------------
TRANSACTIONS
------------
Trx id counter 11271
Purge done for trx's n:o < 11271 undo n:o < 0 state: running but idle
History list length 36
LIST OF TRANSACTIONS FOR EACH SESSION:
---TRANSACTION 421919568688976, not started
0 lock struct(s), heap size 1136, 0 row lock(s)
--------
FILE I/O
--------
I/O thread 0 state: waiting for completed aio requests (insert buffer thread)
I/O thread 1 state: waiting for completed aio requests (log thread)
I/O thread 2 state: waiting for completed aio requests (read thread)

另外，我不想创建一个文件并从php

中读取它

echo file_get_contents( "innodb_status.php" );

Answer 1

使用var_export工作但它包含数组键。

echo "<pre>";
var_export(array_values($innodb_row));
echo "</pre>";

以下工作完美且给出了所需的输出

$remoteconnect = mysqli_connect("host","user","pass","","port","");
    if (!$remoteconnect)
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }
    else {

    $innodb = "show engine innodb status" ;    
    $innodb_result = mysqli_query($remoteconnect,$innodb);

    if (!$innodb_result) {
        printf("Error: %s\n", mysqli_error($remoteconnect));
        exit();          }

    while($innodb_row = mysqli_fetch_array($innodb_result)) {
    echo "<pre>".$innodb_row[0].$innodb_row[1].$innodb_row[2]."</pre>";
    }

         }
    mysqli_close($remoteconnect);

打印mysql命令提示符“\ G”输出在PHP中

1 个答案: