Jump to content

wglenn01

Members
  • Posts

    12
  • Joined

  • Last visited

Posts posted by wglenn01

  1. I'm not getting any errror, everything I try (latest version below) returns nothing. I store that XML response in a variable called $returned, then my latest iteration to try and exctract the "PaymentID" is:

     

    $multi_array = json_decode( json_encode($returned) , 1);

  2. Hi all, need some quick help i'm working on a cart script with Mercury Payments, their "initializePayment" system throws this XML back to me but for some reason I cannot figure out how to parse it to grab the PaymentID out of it. Any clues? I've been playing with SimpleXML for about the last 6 hours before I gave up.


    Thanks a ton!

     

    <?xml version="1.0" encoding="utf-8"?><soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"><soap:Body><InitializePaymentResponse xmlns="http://www.mercurypay.com/"><InitializePaymentResult><ResponseCode>0</ResponseCode><PaymentID>8f19cc14-27fc-40ec-a929-bee4f043387f</PaymentID><Message>Initialize Successful</Message></InitializePaymentResult></InitializePaymentResponse></soap:Body></soap:Envelope>


    William

  3. Well what I mean is, it works as a single search but when I add another, below it to search for a different name, if the first one succeeds and sets $icon, then the second doesn't succeed and sets $icon to empty. Does that make sense?

     

    I have 27 different names I need to loop through to see which one comes back with a match.

  4. Hey all,

    I have a quick question and I know the answer is probably pretty easy but i've been working on it a while now and can't figure out how to get this done.

    SOO the situation is. I am looping through an RSS feed and with each entry I want to search for a brand name within the title, and add a company logo beside the entry.  I was playing with something like this but it doesn't work real well..

     

    // Find Kokatat
    $find = "Kokatat";
    if(strstr($string, $find)) {
    $icon = "koksm.jpg";
    } else {
    	$icon = "";
    }

     

    I appreciate any input!

  5. btw probably need to use the less than or = to / greater than or = to, as if the age matches either end of the span (ie 19 - 34 span - age =19 or 34) it will not show for on  < or > -

     

    make sense?

     

    Will do, thank you all! I think we have it. Two days of scratching my head on that one. Mabey I need to leave the building for a bit ;)

  6. Are you running the query directly in phpMyAdmin? Or can you send your php and html processing query script? Are you echoing all the fields in your php/html?

     

    I'm not sure what you mean. I am running the query on my php page that outputs the results of the query into a select box. I'm just not sure how to get it to select the correct age categores, or if there is a better way to structure the database to make it work.

  7. Ok all I need some help with this. I have been stuck for two days. Basically I have a form that asks a registrant their gender and age, then my script spits out a list of race categories they qualify for. The situation calls for the registrant to be able to register for any category YOUNGER than them, but not older.

     

    Screen Shot #1 shows my database setup (this can be changed if you have any better ideas...) that holds the race category information.

     

    Here is the sql statement I am using to pull a female racer information, where $race_age is their submitted age.

     

    "SELECT * from counts where female=1 and $race_age < age_end and $race_age > age_start"

     

    This works, but doesn't pull any categories YOUNGER Than their age.

     

    I'm stumpped :) Thanks for any help in advance.

     

     

    [attachment deleted by admin]

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.