準備mysqli選擇使用爆炸

在這裏我爆炸上面的字符串與,並得到了陣列$ myArray。準備mysqli選擇使用爆炸

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

但我怎麼能使其成爲一個select語句

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循環

來源

2014-12-25 ABD

無需爆炸。

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

如果您有：使用IN算哪裏的情況下

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

查詢像

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

來源

2014-12-25 06:22:00

謝謝，這是偉大的 – ABD

爲什麼寫多個sql語句，如果你得到了IN運營商，您檢查一些特定的字符串值需要使用這個：

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

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

來源

2014-12-25 06:30:29

我認爲它是一個壞主意，爲每個迭代進行查詢。相反，運行像

$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.")");

來源

2014-12-25 06:32:00 benjosantony

單個查詢您可以使用：

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

或者：

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

來源

2014-12-25 06:33:31

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