利用爆炸制备Mysqli选择

Question

在这里，我用,爆炸了上面的字符串，得到了数组$myArray。

<?php
$myString = '1,2,3,4';
$myArray = explode(',', $myString);
print_r($myArray);

但是如何使它成为select语句？

mysqli_query($con,"SELECT * FROM Persons where id = 'First part of Array'");
mysqli_query($con,"SELECT * FROM Persons where id = 'Second part of Array'");
mysqli_query($con,"SELECT * FROM Persons where id = 'Third part of Array'"); ..

在Foreach循环中

Answer 1

没必要爆炸。使用IN运算符检查何地情况

mysqli_query($con,"SELECT * FROM Persons where id IN (".$myString.")");

类似于的查询

 SELECT * FROM Persons where id IN (1,2,3,4);

Answer 2

如果您有IN运算符，为什么要编写多个sql：

$myArray = explode(',', $myString);

如果您对字符串值有特定的需要，请使用以下命令：

$myArray =array_map('strval', $myArray );

mysqli_query($con,"SELECT * FROM Persons where id IN (".$myString.")");

Answer 3

我认为为每次迭代进行查询是个坏主意。而是运行一个查询，如下所示

$myString = '1,2,3,4';
$myArray = explode(',', $myString);
  // Just format the string so that it appears as  '1','2'... instead of  '1,2'
for ($i = 0; $i < count($myArray); $i ++)
{
   $myArray[$i] = "'" . $myArray[$i] . "'";
}
$myFormattedString = implode(',', $myArray);

mysqli_query($con,"SELECT * FROM Persons where id IN (".$myFormattedString.")");