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php - 在 php 中打印 mysql 命令提示符 "\G"输出

转载 作者:行者123 更新时间:2023-11-29 02:51:47 25 4
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我可以从命令行在文本文件中输出 SHOW ENGINE INNODB STATUS\G;。还有任何使用 mysqli_fetch_array() 或 mysqli_fetch_assoc() 的查询

下面也是可能的,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。

我已经经历了var_dump()serialize()但不能真正让它投入工作。请帮忙

Web 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)

此外,我不想创建一个文件并从 php 读取它

echo file_get_contents( "innodb_status.php" );

最佳答案

使用 var_export 有效,但它包含数组键。

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

下面的代码完美运行并给出了所需的输出

$remoteconnect = mysqli_connect("host","user","pass","","port","");
if (!$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);

关于php - 在 php 中打印 mysql 命令提示符 "\G"输出,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34849311/

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