如何在php

Question

我想echo img每次$i增加echo，但只有$i是最后for($i=0; $i<=5;$i++) $sql="SELECT DISTINCT id, Make FROM sve WHERE id='".$i."'"; $rezult=mysqli_query($db, $sql); $red=mysqli_fetch_object($rezult); echo "<img src='brend/brend$red->id.png' class='imgbrend' strana='".$red->Make."'/>"; 。

make_optional_tuple<int, double, std::string>

Answer 1

您可以使用MySQL的IN或BETWEEN关键字，而不是使用for循环，然后迭代结果集。

您的查询会变成这样：

SELECT DISTINCT id, Make FROM sve WHERE id BETWEEN 0 AND 5

然后你需要一个这样的循环：

while ($red = mysqli_fetch_object($rezult)) {
    echo "<img src='brend/brend" . $red->id . ".png' class='imgbrend' strana='" . $red->Make . "'/>";
}

完整代码：

$sql = "SELECT DISTINCT id, Make FROM sve WHERE id BETWEEN 0 AND 5";

$rezult = mysqli_query($db, $sql);
while ($red = mysqli_fetch_object($rezult)) {
    echo "<img src='brend/brend" . $red->id . ".png' class='imgbrend' strana='" . $red->Make . "'/>";
}