Question

如果有三条记录，我想比较这些值 'test1'和'test 2'并将数据库更新为'open'或'close'，如果三个记录的'test1'和'test2'值匹配相同。但是下面的代码不起作用。

HTML：

    while($row = mysql_fetch_assoc($sql)){
        <input type="text" value="<?php echo $row['test1']; ?>" name="test1[]">
        <input type="text" value="" name="test2[]">
    }
  <input name="Submit" type="submit" value="update">

PHP：

<?php

if (isset($_POST['Submit']))
{
$test1=$_POST['test1'];
$test2=$_POST['test2'];

if($test1==$test2)
{
$sql1="UPDATE table SET status='close' WHERE test1='$test1' AND test2='$test2' ";
$db=mysql_query($sql1);
}
else
{
$sql1="UPDATE table SET status='open' WHERE test1='$test1' AND test2='$test2' ";
$db=mysql_query($sql1);
}
}

?>

Answer 1

您需要比较数组值。尝试下面的代码，

<?php
    $three = array_intersect($test1,$test2);

    if(count($three) > 2)
    {
    $sql1="UPDATE table SET status='close' WHERE test1='$test1' AND test2='$test2' ";
    $db=mysql_query($sql1);
    }
    else
    {
    $sql1="UPDATE table SET status='open' WHERE test1='$test1' AND test2='$test2' ";
    $db=mysql_query($sql1);
    }
?>

比较多个数组值并更新表状态

1 个答案: