Echo different images depending on array value

Question

I want to echo a different image depending on if a key exists or not.

Here is an example of the array I'm using

["Person"] => array(11) {
    ["id"] => int(38482818123)
    ["weight"] => int(140)
    ["height"] => int(65)
}
["Name"] => array(2) {
    ["firstname"] => string(4) "John"
    ["lastname"] => string(5) "Smith"
}

So the name field isn't always there. I need to showimage a if the name is there and image b if there is no name.

What I've tried:

foreach($personArray as $person)
{
    if ($person['Name'] != '')
    {
        echo "<img src='image-a.png'>";
    }
    else
    {
        echo "<img src='image-b.png'>";
    }
}

Now the problem I have is that even though the person has a name, I'm seeing both images on the page instead of just image a

I have also tried using array_key_exists("Name", $personArray); but for some reason I am getting bool(false) as a result.

You're iterating an array. So if some people have names and some don't, you'll probably see multiple copies of each image. — mkasberg
– mkasberg, Commented May 1, 2016 at 0:09
empty() might be the function you're looking for. !empty($person['Name']) — mkasberg
– mkasberg, Commented May 1, 2016 at 0:10
From what I can see, the Person array does not contain a name value to begin with. Name is a separate array altogether? — CmdrSharp
– CmdrSharp, Commented May 1, 2016 at 0:13
That's a typo on my part I believe, I updated the question with an image of the array as I see it. — Mr.Smithyyy
– Mr.Smithyyy, Commented May 1, 2016 at 0:43

Captain Red · Accepted Answer · 2016-05-02 13:33:25Z

3

First, it seems that you have a different array for the Name, I wonder why you don't have something like:

 ["Person"] => array(11) {
    ["id"] => int(38482818123)
    ["weight"] => int(140)
    ["height"] => int(65)
    ["firstname"] => string(4) "John"
    ["lastname"] => string(5) "Smith"
    }

As per that case, the array index with the key "firstname" , "lastname" might not be set for some person. So you can check if that index is set using

isset(); function

Try:

foreach($personArray as $person)
{
    if (isset($person['firstname']) && isset($person['lastname']) ) // You may use or as well ||
    {
        echo "<img src='image-a.png'>";
    }
    else
    {
        echo "<img src='image-b.png'>";
    }


   //  Or as a short hand if statement:

  echo (isset($person['firstname'])&& isset($person['lastname'])) ? "<img src='image-a.png'>" : "<img src='image-b.png'>";
}

Edit: 1 You still seem to be using two different arrays: Person and Name. Doing this, you cannot say which particular person has the name values or not: For eg: if you have 10 persons and have the Name array with firstname, lastname for just 7 person then you will have person[0], person[1].....person[9] and Name[0]...Name[6]. As per your construct, there is no any reference / link between the Person and Name array. Suppose 1st person has Name, then Person[0] and Name[0] would represent the same person. But if the first 3 Person do not have Name, then Person[3] will have Name[0]...and so on, hence, it is not possible to identify to which Person does a particular Name array belong to. And Note: you cannot use Name["firstname"] inside the foreach() of Person. Because, your name array would be of the form Name[0]["firstname"], Name[0]["lastname"] and so on.

Bottom Line:

If possible, try to use / include the firstname, lastname in the Person array itself. That way, when iterating / looping through the Person array using foreach() you can check if each of these person have firstname , lastname or not: Hope this is clear.

edited May 2, 2016 at 13:33

answered May 1, 2016 at 0:14

Captain Red

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3 Comments

Mr.Smithyyy Over a year ago

I tried your answer but I still got both images. I might have portrayed the array poorly in my question but I took a picture of it if you would like to take another look.

Mr.Smithyyy Over a year ago

Your code also works with my array outside of the loop, thank you!

Captain Red Over a year ago

Well great then :)

Death-is-the-real-truth · Accepted Answer · 2016-05-01 00:47:38Z

1

You need to use empty() like below:-

 if(!empty($personArray['person']['Name'])){ 
    //image a code
 }else{
    //image b code
 }

edited May 1, 2016 at 0:47

answered May 1, 2016 at 0:14

Death-is-the-real-truth

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4 Comments

Mr.Smithyyy Over a year ago

It's still showing both, I don't understand why it is doing that for an array with 1 person.

Mr.Smithyyy Over a year ago

Can I use this outside of the loop?

Mr.Smithyyy Over a year ago

my array is json_decoded inside a variable named $result. I just tried your solution outside of my loop and it gives image 2 if the name field is filled in.

Mr.Smithyyy Over a year ago

Let me take a picture of my exact array and post it in my question, I'm still getting the second image when the field is filled in. I might be portraying it wrong in my question.

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