-1

あるクエリで郵便番号を取得し、それを別のクエリの where 句で使用しようとしています。

<?php
$mobile = '07790807055';
//$mobile = $_POST['mobile'];


mysql_connect("", "", "") or die(mysql_error());
mysql_select_db("") or die(mysql_error());

$query =("SELECT POSTCODE FROM appregistration WHERE MOBILE_NUMBER = '$mobile'");
$result = mysql_query ($query) or die ("Unable to connect. " . mysql_error());
$row = mysql_fetch_array($result);
$postcode = $row['POSTCODE'];
/// gets postcode from appregistration table 

$sql=mysql_query("SELECT INCIDENT_ID, INVESTIGATION,TYPE_OF_INCIDENT,DESCRIPTION FROM appreports WHERE POSTCODE = '$postcode'");
//uses postcode from ppregistration table  to find info from appreports table 

while($row=mysql_fetch_assoc($sql)) $output[]=$row;

print(json_encode($output));

mysql_close();
?>
4

2 に答える 2

2

JOIN2 つのクエリを実行するのではなく、a を使用しない理由:

SELECT `INCIDENT_ID`, `INVESTIGATION`, `TYPE_OF_INCIDENT`, `DESCRIPTION`
FROM `appreports`
INNER JOIN `appregistration` ON `appreports`.`POSTCODE` = `appregistration`.`POSTCODE`
WHERE `appregistration`.`MOBILE_NUMBER` = '$mobile'

例えば：

<?php
$mobile = '07790807055';

mysql_connect("", "", "") or die(mysql_error());
mysql_select_db("") or die(mysql_error());

$sql = mysql_query("SELECT `INCIDENT_ID`, `INVESTIGATION`, `TYPE_OF_INCIDENT`, `DESCRIPTION`
                    FROM `appreports`
                    INNER JOIN `appregistration` 
                    ON `appreports`.`POSTCODE` = `appregistration`.`POSTCODE`
                    WHERE `appregistration`.`MOBILE_NUMBER` = '$mobile'");

while($row=mysql_fetch_assoc($sql))
{
    $output[] = $row;
}

print(json_encode($output));

mysql_close();
?>
于 2013-03-05T20:50:01.927 に答える