Question

我看过许多不同的网站，包括php.net，但仍然没有找到我的插入脚本无效的原因。我正在使用Eclipse在PHP中编程，我安装了PHP插件，以便它能正常工作，但我还没弄清楚我的脚本有什么问题。

实施例。

switch($type) {

    case(1):
        require("opendb.php");
        $query = "INSERT INTO ability(AbilityName,Description,Strength,Defense,Luck,Agility) values ('" . $name . '","' . $description . '","' . $strength . '","' . $defense . '","' . $luck . '","' . $agility . "')";
        if (!mysql_query($query)) {
            echo "Ability was not entered successfully: " . mysql_error();
        } else {
            echo "Ability was entered successfully!";
        }
        break;
    case(2):
        require("opendb.php");
        $query = "INSERT INTO weapon(WeaponName,Description,Strength,Defense,Luck,Agility) values ('"  . $name . '","' . $description . '","' . $strength . '","' . $defense . '","' . $luck . '","' . $agility . "')";
        if (!mysql_query($query)) {
            echo "Weapon was not entered successfully: " . nysql_error();
        } else {
            echo "Weapon was entered successfully!";
        }
        break;
    default:
        require("opendb.php");
        $query = "INSERT INTO item(ItemName,Description,Strength,Defense,Luck,Agility) values ('" . $name . '","' . $description . '","' . $strength . '","' . $defense . '","' . $luck . '","' . $agility . "')";
        if (!mysql_query($query)) {
            echo "Item was not entered successfully: " . mysql_error();
        } else {
            echo "Item was entered successfully!";
        }
    }
}

您可以选择单选按钮将对象作为项目，武器或能力插入，但我一直得到列值与行值计数错误不匹配。这是我创建的三个表。每次输入对象时，ID都会自动递增。它不是null，但我不需要在我用来输入对象的表单上包含ID。

项目表：

ItemID INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
ItemName varchar(40),
Description longtext,
Strength INT NOT NULL,
Defense INT NOT NULL,
Luck INT NOT NULL,
Agility INT NOT NULL

武器表：

WeaponID INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
WeaponName varchar(40),
Description longtext,
Strength INT NOT NULL,
Defense INT NOT NULL,
Luck INT NOT NULL,
Agility INT NOT NULL

能力表：

AbilityID INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
AbilityName varchar(40),
Description longtext,
Strength INT NOT NULL,
Defense INT NOT NULL,
Luck INT NOT NULL,
Agility INT NOT NULL

Answer 1

您使用的是无效的字符串引用，这是SQL命令中的一个简单语法错误。

如果您回显此行，

$query = "INSERT INTO ability(AbilityName,Description,Strength,Defense,Luck,Agility) values ('" . $name . '","' . $description . '","' . $strength . '","' . $defense . '","' . $luck . '","' . $agility . "')";

输出将如下：

INSERT INTO ability(AbilityName,Description,Strength,Defense,Luck,Agility) values ('name","description","strength","defense","luck","agility')

如您所见，第一列和最后一列使用混合字符串引号，这会导致错误。我建议你在SQL查询中只使用单引号，并使用HEREDOC方法在PHP中创建这些字符串。

如何修复列数与行值计数不匹配

1 个答案: