[英]How do I get the result into a variable MYSQL
I have the following PHP script. 我有以下PHP脚本。 How do I get the correct results.
我如何获得正确的结果。 Each query should only return a number.
每个查询应仅返回一个数字。 I need to get
rank
in this case. 在这种情况下,我需要获得
rank
。 I was also wondering how to combine those two queries into one statement. 我还想知道如何将这两个查询合并为一个语句。
function getRankings($country, $deviceid)
{
$queryWorld = "SELECT 1 + (SELECT count( * ) FROM ScoreTable a WHERE a.score > b.score ) AS rank FROM
ScoreTable b WHERE DeviceID='$deviceid' ORDER BY rank LIMIT 1";
$queryCountry = "SELECT 1 + (SELECT count( * ) FROM ScoreTable a WHERE a.score > b.score AND Country='$country') AS rank FROM ScoreTable b WHERE DeviceID='$deviceid' ORDER BY rank LIMIT 1";
$resultWorld = mysql_query($queryWorld) or die(mysql_error());
$rowWorld = mysql_fetch_row($resultWorld);
$resultCountry = mysql_query($queryCountry) or die(mysql_error());
$rowCountry = mysql_fetch_row($resultCountry);
$arr = array();
$arr[] = array("WorldRanking" => $rowWorld[0], "CountryRanking" => $rowCountry[0]);
echo json_encode($arr);
}
If I type the queries individually into MYSQl I get the correct answers. 如果我将查询分别键入MYSQl,则会得到正确的答案。 But the
echo
produces 但是
echo
产生
[{"WorldRanking":null,"CountryRanking":null}]
It should be something like 应该是这样的
[{"WorldRanking":"4","CountryRanking":"1"}]
I think I need to get the value of rank
but I do not know how. 我想我需要获得
rank
的价值,但我不知道如何。
try your code like this, but I don't know this code correct or not: 尝试这样的代码,但我不知道此代码正确与否:
function getRankings($country, $deviceid)
{
$queryWorld = "SELECT 1 + (SELECT count( * ) FROM ScoreTable a WHERE a.score > b.score ) AS rank FROM
ScoreTable b WHERE DeviceID='$deviceid' ORDER BY rank LIMIT 1";
$queryCountry = "SELECT 1 + (SELECT count( * ) FROM ScoreTable a WHERE a.score > b.score AND Country='$country') AS rank FROM ScoreTable b WHERE DeviceID='$deviceid' ORDER BY rank LIMIT 1";
$resultWorld = mysql_query($queryWorld) or die(mysql_error());
while ($rowWorld = mysql_fetch_row($resultWorld)){
$value1 = $rowWorld['filedname1'];
}
$resultCountry = mysql_query($queryCountry) or die(mysql_error());
while($rowCountry = mysql_fetch_row($resultCountry)){
$value2 = $rowWorld['filedname2'];
}
$arr[] = array("WorldRanking" => $value1, "CountryRanking" => $value2);
print_r($arr);
}
<?php
function getRankings($country, $deviceid)
{
$queryWorld = "SELECT 1 + (SELECT count( * ) FROM ScoreTable a WHERE a.score > b.score ) AS rank FROM
ScoreTable b WHERE DeviceID=$deviceid ORDER BY rank LIMIT 1";
$queryCountry = "SELECT 1 + (SELECT count( * ) FROM ScoreTable a WHERE a.score > b.score AND Country='$country') AS rank FROM ScoreTable b WHERE DeviceID=$deviceid ORDER BY rank LIMIT 1";
$resultWorld = mysql_query($queryWorld) or die(mysql_error());
$no_of_results_in_resultWorld = mysql_num_rows($resultWorld);
if($no_of_results_in_resultWorld > 0){
$rowWorld = mysql_fetch_row($resultWorld);
}
$resultCountry = mysql_query($queryCountry) or die(mysql_error());
$no_of_results_in_resultCountry = mysql_num_rows($resultCountry);
if($no_of_results_in_resultCountry > 0){
$rowCountry = mysql_fetch_row($resultCountry);
}
$arr = array();
$arr[] = array("WorldRanking" => $rowWorld[0], "CountryRanking" => $rowCountry[0]);
echo json_encode($arr);
}
check whether it is returning results or not. 检查它是否返回结果。
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