mysql在PHP中选择不同的查询 [英] mysql select distinct query in PHP

问题描述

$sql = "SELECT DISTINCT Branch FROM student_main";
    $result = mysql_query($sql);
    $row_num = mysql_num_rows($result);
    $rows = mysql_fetch_array($result);
    echo "<select name='Branch'>";
    for($i=0;$i<=$row_num-1;$i++){
        echo "<option value='".$rows[$i]."'>".$rows[$i]."</option>";

    }
    echo "</select>";
    echo "<input type='submit' Value='submit' />";
    echo "</form>";

我正在尝试使用上述代码为我的表单创建一个下拉列表。但它不工作。 分支列中有3个不同的值，但在下拉列表中，它只显示一个值（第一个），而后两个值作为空值。

I am trying to create a dropdown using the above code for my form. But its not working. There are 3 distinct values in the Branch column but in the dropdown, it shows only one value(the first one) and the next two as blank values.

但是在echo $ row_num中，它显示为3.

这意味着它提取三行，但为什么它不显示在下拉列表中名单。

However when in echo $row_num, its shows 3.
Thats means its fetching the three rows, but then why its not showing in the dropdown list.

如果我在phpmyadmin中运行相同的查询，它会显示正确的答案，它会返回3个不同的分支值。

If I run the same query in phpmyadmin it shows the correct answer i.r it returns 3 distinct Branch values.

推荐答案

你应该这样做：

$sql = "SELECT DISTINCT Branch FROM student_main";
$result = mysql_query($sql);

echo "<select name='Branch'>";
while ($row = mysql_fetch_array($result)) {
    echo "<option value='".$row[0]."'>".$row[0]."</option>";
}
echo "</select>";

echo "<input type='submit' Value='submit' />";
echo "</form>";

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