Question

我需要更新单个数据库中单个表中的多个列。现在，当我运行PHP脚本时，它仅完成第一个UPDATE命令，而不会更新其后的条目。我不确定这是怎么回事：

sudo service nginx restart

当我从命令行运行它时，它不会出错，它确实可以运行第一个更新。当我检查数据库时，实际上它已将ID为“ 10”的“ topic_id”列中的“ team_id”更新为“ 15”，但其他两列未更新。为什么会这样？

Answer 1

您要一遍又一遍地定义相同的变量，从而踩踏以前的版本。仅最后一个定义会“保留”并运行。

切换到数组：

// Define an array
$sql = [ ];
// Append to the array each query
$sql[] = "UPDATE ost_help_topic SET team_id='15' WHERE topic_id=10";
$sql[] = "UPDATE ost_help_topic SET team_id='15' WHERE topic_id=17";
$sql[] = "UPDATE ost_help_topic SET team_id='15' WHERE topic_id=22";

foreach($sql as $q) {
  // Run query.
}

或使用IN (...)方法：

UPDATE ost_help_topic SET team_id='15' WHERE topic_id IN (10,17,22)

其中一个查询可以将多个值更新为相同。

Answer 2

您需要执行每一行：

//1
$sql="UPDATE ost_help_topic SET team_id='15' WHERE topic_id=10";

if (mysqli_query($conn, $sql)) {
        echo "Record updated successfully";
} else {
        echo "Error updating record: " . mysqli_error($conn);
}

//2
$sql="UPDATE ost_help_topic SET team_id='15' WHERE topic_id=17";

if (mysqli_query($conn, $sql)) {
        echo "Record updated successfully";
} else {
        echo "Error updating record: " . mysqli_error($conn);
}

//2
$sql="UPDATE ost_help_topic SET team_id='15' WHERE topic_id=22";

if (mysqli_query($conn, $sql)) {
        echo "Record updated successfully";
} else {
        echo "Error updating record: " . mysqli_error($conn);
}

或创建一个数组

$arra = array();
$arra[]="UPDATE ost_help_topic SET team_id='15' WHERE topic_id=10";
$arra[]="UPDATE ost_help_topic SET team_id='11' WHERE topic_id=12";
$arra[]="UPDATE ost_help_topic SET team_id='12' WHERE topic_id=13";

和foreach：

foreach ($arra as $sql) {

   if (mysqli_query($conn, $sql)) {
            echo "Record updated successfully";
    } else {
            echo "Error updating record: " . mysqli_error($conn);
    }
}

Answer 3

仅使用“或”运算符即可解决问题！

$sql = "UPDATE ost_help_topic SET team_id='15' WHERE topic_id=10 OR topic_id=12 OR topic_id=13"

PHP脚本更新MySQL表中的多个列

3 个答案: