mysql从表中获取行值，其中列值等于特定值

Question

我有一个表格，其中modeltest的结果按降序排列在列中。现在我想获取特定行的行号。以下是示例表：

所以当我用userid = 44搜索a时，它应该返回3.再次当我用userid = 11搜索时，它应该返回4.我怎么能这样做？如果有人可以帮助我。

Answer 1

一个简单的where子句将实现这一点。

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        if (height === 0 && width > 0)
            height = width / this.data.drawn_units.length;
        else if (width === 0 && height > 0)
            width = height * this.data.drawn_units.length;
        
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        this.data.attributes.item_size = Math.min(width / this.data.drawn_units.length, height);
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        this.data.attributes.radius = ((this.data.attributes.item_size * 0.8) - this.data.attributes.line_width) / 2;
        this.data.attributes.outer_radius = this.data.attributes.radius + 0.5 * Math.max(this.data.attributes.line_width, this.data.attributes.line_width * this.config.bg_width);

将表和userid值交换为所需的值。

希望这有帮助

Answer 2

例如：

$stmt = PDO::prepare('SELECT id FROM table WHERE userid = ?');
$userId = 44;

$stmt->execute([$userId]);
$result = $stmt->fetch(PDO::FETCH_OBJ);

echo $result->id;

http://php.net/manual/en/book.pdo.php