PHP script returning JSON not working

Question

The below returns "\nQuery was empty" as I run it simply from my server with the URL in the Browser.

This is the PHP code:

    <?

$databasehost = "server";
$databasename = "xxxx";
$databaseusername ="xxxx";
$databasepassword = "xxxx";
$query = "SELECT * FROM  `Tailor`LIMIT 0 , 30";
$con = mysql_connect($databasehost,$databaseusername,$databasepassword) or die(mysql_error());
mysql_select_db($databasename) or die(mysql_error());
$query = file_get_contents("php://input"); 
$sth = mysql_query($query);

if (mysql_errno()) { 
    header("HTTP/1.1 500 Internal Server Error");
    echo $query.'\n';
    echo mysql_error(); 
}
else
{
    $rows = array();
    while($r = mysql_fetch_assoc($sth)) {
        $rows[] = $r;
    }
    print json_encode($rows);
}
?>

Kishan Patel · Accepted Answer · 2013-05-02 05:56:36Z

1

No need to use file_get_content and you have to put one white space after table name.

<?php

$databasehost = "server";
$databasename = "xxxx";
$databaseusername ="xxxx";
$databasepassword = "xxxx";
$query = "SELECT * FROM  `Tailor` LIMIT 0 , 30";
$con = mysql_connect($databasehost,$databaseusername,$databasepassword) or die(mysql_error());
 mysql_select_db($databasename) or die(mysql_error());

$sth = mysql_query($query);

if (mysql_errno()) { 
header("HTTP/1.1 500 Internal Server Error");
echo $query.'\n';
echo mysql_error(); 
}
else
{
$rows = array();
while($r = mysql_fetch_assoc($sth)) {
    $rows[] = $r;
}
print json_encode($rows);
}
?>

answered May 2, 2013 at 5:56

Kishan Patel

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Comments

chandresh_cool · Accepted Answer · 2013-05-02 05:51:38Z

0

Your query has a problem you need whitespace after tablename

SELECT * FROM  `Tailor` LIMIT 0 , 30

answered May 2, 2013 at 5:51

chandresh_cool

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4 Comments

Naaz Over a year ago

cured it BUT still the same

chandresh_cool Over a year ago

WHy are you sing this $query = file_get_contents("php://input");?

Naaz Over a year ago

i removed it was not needed nw im watching out for result

chandresh_cool Over a year ago

Okay let me know the outcome

JOE LEE · Accepted Answer · 2013-05-02 06:01:48Z

0

first

$query = "SELECT * FROM  `Tailor`LIMIT 0 , 30";

but you overwrite it

$query = file_get_contents("php://input"); 

// $query like   aaa=ggg&bbb=kkk

then you

$sth = mysql_query($query);

// LIKE $sth = mysql_query('aaa=bbb&ccc=lll'); //it not sql query format

answered May 2, 2013 at 6:01

JOE LEE

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