坐标搜索

Question

现在差不多可以用了。我可以在表格中得到结果，但由于某种原因，它们不太正确。

搜索坐标x 100 &y 100

我得到x 1-25 &只有Y 1-25而不是x 74-125 &74 - 125 y

<?php
$x = $_POST['x'];
$y = $_POST['y'];
mysql_connect ("localhost","user","pass")  or die (mysql_error());
mysql_select_db ("db");

$res = mysql_query("select * FROM table WHERE (x between $x-25 AND $x+25) AND (y BETWEEN $y-25 AND $y+25)");
 echo "<table border='1' align='center' cellpadding='5'>";
    echo "<tr> <th>City Name</th> <th>X</th> <th>Y</th> </tr>";
    // loop through results of database query, displaying them in the table
    while($row = mysql_fetch_array( $res )) {
            // echo out the contents of each row into a table
echo "<tr>";
echo '<td>' . $row['cityname'] . '</td>';
    echo '<td>' . $row['x'] . '</td>';
echo '<td>' . $row['y'] . '</td>';
    echo "</tr>";
    } 
      // close table>
    echo "</table>";
?>

Answer 1

试试这个:

$query="SELECT * FROM cityname
WHERE (x between".$_POST['search']."-25 AND ".$_POST['search']."+25)
AND (y between".$_POST['search']."-25 AND ".$_POST['search']."+25)";

Answer 2

当我告诉你这些的时候，请记住，以你现在的方式来做会让你更容易受到SQL注入的影响，并且你应该真正使用PDO或类似的方法来处理准备好的语句。话虽如此:

$x = (intval)$_POST['fld_x'];
$y = (intval)$_POST['fld_y'];
$res = mysql_query("select * FROM my_dbase WHERE (x between $x-25 AND $x+25) AND (y BETWEEN $y-25 AND $y+25)");
$data = array();
while($row = mysql_fetch_assoc($res)) {
    $data[] = $row;
}

然后做你想做的$data

示例

编辑

为确保数据库连接正常，请在需要的地方更改/添加以下内容:

if(!mysql_select_db ("dbase")) {
    die(mysql_error());
}

after after $res = mysql_query... put:

if(!$res) {
    die("Query Failed: ".mysql_error());
}

编辑

基于当前代码的完整代码:

前面显示的HTML为:

<form action="xysearch.php" method="get">
<label>X Coord
<input type="text" name="fld_x" />
</label>
<label>Y Coord
<input type="text" name=fld_y" />
</label>
<input type="submit" value="Search" />
</form>

， PHP是:

<?php
$x = (intval)$_POST['fld_x'];
$y = (intval)$_POST['fld_y'];
mysql_connect ("localhost","user","pass")  or die (mysql_error());
mysql_select_db ("db");

$res = mysql_query("select * FROM table WHERE (x between $x-25 AND $x+25) AND (y BETWEEN $y-25 AND $y+25)");
 echo "<table border='1' align='center' cellpadding='5'>";
 echo "<tr> <th>City Name</th> <th>X</th> <th>Y</th> </tr>";
// loop through results of database query, displaying them in the table
while($row = mysql_fetch_array( $res )) {
    // echo out the contents of each row into a table
    echo "<tr>";
    echo '<td>' . $row['cityname'] . '</td>';
    echo '<td>' . $row['x'] . '</td>';
    echo '<td>' . $row['y'] . '</td>';
    echo "</tr>";
} 
// close table>
echo "</table>";
?>

Answer 3

如果你想让圆的半径为25，你需要使用毕达哥拉斯定理

所以查询应该是

SELECT cityname, x, y FROM my_database
WHERE SQRT((x - centre_x) * (x - centre_x) + (y - centre_y) * (y - centre_y)) <=25

其中centre_x, centre_y是您输入的位置