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DeclanFell
DeclanFell

Reputation: 11

PHP Not changing color when equals the value

I have a quick question, I'm not sure what I have done wrong in this part of the code but every time Type(NUM) equals the value of the if statement.

I've tried with 2 pieces of data

ID 1 = Type(NUM) 2

ID 2 = Type(NUM) 1

But it's only going to the first if statement and stopping.

Picture of this (https://i.sstatic.net/Of4B6.png)

The code is;

    while($row = mysqli_fetch_assoc($result)){
    if($row["Commit"]== "test"){
        continue;
    }
    echo "<tr ID = \"cell\">";
    if($row['Type(NUM)'] = 1){

        echo '<td style="background-color:#ffab0a; font-weight:bold"> UPDATE</td>';
    }
    else if($row['Type(NUM)'] = 2){

        echo '<td style="background-color:#00cc00; font-weight:bold">NEW </td>';
    }
        else if($row['Type(NUM)'] =3){

        echo '<td style="background-color:#ff00ff; font-weight:bold"> FIX </td>';
    }
            else if($row['Type(NUM)'] =4 ){

        echo '<td style="background-color:#cc0000; font-weight:bold"> REMOVE </td>';
    }
    echo "<td>".$row['ID']."</td>";

Upvotes: 0

Views: 30

Answers (2)

Frank Leal
Frank Leal

Reputation: 206

you every time do the if you are making the $var to num use ' == ' instead of ' = '

Upvotes: 0

Ahmed Ebead
Ahmed Ebead

Reputation: 50

U must use $row['Type(NUM)'] == 2 not '='

while($row = mysqli_fetch_assoc($result)){
if($row["Commit"]== "test"){
    continue;
}
echo "<tr ID = \"cell\">";
if($row['Type(NUM)'] == 1){

    echo '<td style="background-color:#ffab0a; font-weight:bold"> UPDATE</td>';
}
else if($row['Type(NUM)'] == 2){

    echo '<td style="background-color:#00cc00; font-weight:bold">NEW </td>';
}
    else if($row['Type(NUM)'] ==3){

    echo '<td style="background-color:#ff00ff; font-weight:bold"> FIX </td>';
}
        else if($row['Type(NUM)'] ==4 ){

    echo '<td style="background-color:#cc0000; font-weight:bold"> REMOVE </td>';
}
echo "<td>".$row['ID']."</td>";

Upvotes: 1

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