Question

在这里，我用,分解了上面的字符串，得到了数组$ myArray。

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

但我怎样才能使它成为一个选择陈述

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

无需爆炸。使用IN运算符

签入案例

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

查询

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

Answer 2

如果您携带 IN 运算符，为什么要写多个sqls：

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

如果您对字符串值有特定需求，请使用：

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

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

Answer 3

我认为对每次迭代进行查询都是个坏主意。而是运行像

这样的单个查询

$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.")");

Answer 4

您可以使用：

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

或者：

$ids = join(',',$myArray); 
mysqli_query($con,"SELECT * FROM Persons where IN ($ids)");

准备Mysqli选择使用爆炸

4 个答案: