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webdeveloper123

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Posts posted by webdeveloper123

    using for loop to display elements from array

    in PHP Coding Help

    Posted

    Hey Barand, yes I used foreach before on this array sucessfully but then came across this for loop and thought it look interesting and so I gave it a try.

    I'm Using: 

    foreach ($table as $row => $data) {
  echo "<p><b>Row number $row</b></p>";
  echo "<ul>";
  foreach ($data as $col) {
    echo "<li>$col[fname]</li>";
    echo "<li>$col[lname]</li>";
    echo "<li>$col[age]</li>";
  }
  echo "</ul>";
}

    and I am getting the data from $newdata but getting illegal offset errors all over the page for the other array. Plus the data from $newdata shows all records under Row Number 4 but apparently they should all be in 1 array now ($table)

    using for loop to display elements from array

    in PHP Coding Help

    Posted

    Hi,

    I've got a for loop  which displays elements from an array. The array looks like this: 

    $table = array ( 
            0 => array ( 
                'fname' => 'Peter', 
                'lname' => 'Smith', 
                'age' => '37'
                ), 
            1 => array ( 
                'fname' => 'Paul', 
                'lname' => 'Hartley', 
                'age' => '48'
                ), 
            2 => array ( 
                'fname' => 'Mary', 
                'lname' => 'Baker', 
                'age' => '42'
                ), 
            3 => array ( 
                'fname' => 'Jane', 
                'lname' => 'Doe', 
                'age' => '51'
              
                )
            );

    The for loop looks like this: 

    for ($row = 0; $row < 4; $row++) {
  echo "<p><b>Row number $row</b></p>";
  echo "<ul>";
  for ($col = 0; $col < 1; $col++) {
    echo "<li>".$table[$row]['fname']."</li>";
     echo "<li>".$table[$row]['lname']."</li>";
      echo "<li>".$table[$row]['age']."</li>";
  }
  echo "</ul>";
}

    Now this code is working fine. Then I set about adding an element to the array using array_push like this: 

    $newdata =  array (
      'fname' => 'Lionel',
      'lname' => 'Messi',
      'age' => '36'
    );



array_push($table,$newdata);

    And I displayed it (sucessfully) using this: 

    for ($row = 0; $row < 5; $row++) {
  echo "<p><b>Row number $row</b></p>";
  echo "<ul>";
  for ($col = 0; $col < 1; $col++) {
    echo "<li>".$table[$row]['fname']."</li>";
     echo "<li>".$table[$row]['lname']."</li>";
      echo "<li>".$table[$row]['age']."</li>";
  }
  echo "</ul>";
}

    Note: The only difference between the above 2 for loops is I changed $row < 4 to $row < 5

    Then I wanted to add more than one element to the array so I used this: 

    $newdata = array ( 
            4 => array ( 
                'fname' => 'Jon', 
                'lname' => 'Atkins', 
                'age' => '27'
                ), 
            5 => array ( 
                'fname' => 'Phil', 
                'lname' => 'Jones', 
                'age' => '14'
                ), 
            6 => array ( 
                'fname' => 'Frank', 
                'lname' => 'Lampard', 
                'age' => '48'
                ), 
            7 => array ( 
                'fname' => 'Toney', 
                'lname' => 'Brentford', 
                'age' => '25'
              
                )
            );

array_push($table,$newdata);

    But the weird baffling part is I am using the same forloop: 

    for ($row = 0; $row < 8; $row++) {
  echo "<p><b>Row number $row</b></p>";
  echo "<ul>";
  for ($col = 0; $col < 1; $col++) {
    echo "<li>".$table[$row]['fname']."</li>";
     echo "<li>".$table[$row]['lname']."</li>";
      echo "<li>".$table[$row]['age']."</li>";
  }
  echo "</ul>";
}

    Note : This time I have changed $row < 8 but I get this error : 

    Row number 4


Notice: Undefined index: fname
 Undefined index: lname
Undefined index: age

    And errors like : 

    Row number 5


Notice: Undefined offset: 5 in on line 86

Notice: Trying to access array offset on value of type null in  on line 86

Notice: Undefined offset: 5 in  on line 87

Notice: Trying to access array offset on value of type null in  on line 87

Notice: Undefined offset: 5 in  on line 88

Notice: Trying to access array offset on value of type null in  on line 88

    And I get the same errors as above for row 5 but for rows 6 and 7.

    Btw, I have row numbers for each element in the array. So Row Number 0 displays the First Name, last name and age for the element at index 0 etc.

    And I have done a print_r($table) and get this: 

    Array ( [0] => Array ( [fname] => Peter [lname] => Smith [age] => 37 ) [1] => Array ( [fname] => Paul [lname] => Hartley [age] => 48 ) [2] => Array ( [fname] => Mary [lname] => Baker [age] => 42 ) [3] => Array ( [fname] => Jane [lname] => Doe [age] => 51 ) [4] => Array ( [4] => Array ( [fname] => Jon [lname] => Atkins [age] => 27 ) [5] => Array ( [fname] => Phil [lname] => Jones [age] => 14 ) [6] => Array ( [fname] => Frank [lname] => Lampard [age] => 48 ) [7] => Array ( [fname] => Toney [lname] => Brentford [age] => 25 ) ) )

    It seems there are two [4] in the array. I would ideally like to get the rows automatically adjusted so I don't have to keep changing the code when I add new elements in the array. I used  

    $arrayLength = count($table);

    And changed it to: (only showing relevant code here, the rest is same as above for loops): 

    $row < $arrayLength;

    and I get these errors: 

    Notice: Undefined index: fname  on line 86

Notice: Undefined index: lname   on line 87

Notice: Undefined index: age  on line 88

    But the offset errors disappear 

    Sorry for long post

    Many thanks

    creating a shopping list and then delete item with double click

    in Javascript Help

    Posted

    figured it out. anyone who is interested it's this:
      

           document.getElementById("addNew").onclick = function () {
            addOne();
        }
           document.getElementById("sList").ondblclick = function () {
            deleteList();
        }
        function addOne() {
            var a = document.getElementById("addItem").value;
            var li = document.createElement("li");
            li.appendChild(document.createTextNode(a));
            document.getElementById("sList").appendChild(li);

        }
          function deleteList() {
           const list = document.getElementById("sList");
  list.removeChild(list.firstElementChild);
        }

    and HTML same as above

    creating a shopping list and then delete item with double click

    in Javascript Help

    Posted

    Hi,

    I am fairly new to JS and need some advice/logic. I am trying to create a Shopping list using DOM, which adds an item to a list and then when I want to remove it, it should remove on double click. Problem is when I double-click, it deletes all the items, not just the one I double clicked on. I'm pretty sure I have to generate an array of DOM elements and go from there, But I am stuck as what to do next. Here is my code: 

      document.getElementById("addNew").onclick = function () {
            addOne();
        }
          document.getElementById("sList").ondblclick = function () {
            deleteList();
        }
        function addOne() {
            var a = document.getElementById("addItem").value;
            var li = document.createElement("li");
            li.appendChild(document.createTextNode(a));
            document.getElementById("sList").appendChild(li);

        }
        function deleteList() {
             const element = document.getElementById("sList");
            element.remove();
        }

    And the HTML: 

        <div id="message">JavaScript Shopping list</div>
    <div>
        <input type="text" id="addItem">
        <input type="button" id="addNew" value="Add to List"> </div>
    <div id="output">
        <h1>Shopping List</h1>
        <ol id="sList"> </ol>
    </div>

    Many thanks

    echo javascript inside anchor element

    in PHP Coding Help

    Posted

    Hey, sorry guys I know this is a easy one but I've been at it all day and can't seem to figure it out.

    I want to include this: 

    onclick="return confirm('Are you sure?')"

    Inside this line of code: 

    <td> <?php echo("<a href='delete.php?user_id=" . $d_row["FormId"] . "'>Delete</a>"); ?></td>

    I am deleting a record, I have all the SQL, it's all fine, but just this annoying little problem. I have tried many things , including escaping the double quotes but I have got no where. 

    I have tried the following but none of them work: 

    <td> <?php echo("<a href='delete.php?user_id=" . $d_row["FormId"] . "' onclick="return confirm('Are you sure?')">Delete</a>"); ?></td>

  <td> <?php echo("<a href='delete.php?user_id=" . $d_row["FormId"] . "onclick=\"return confirm('Are you sure?')\"'>Delete</a>"); ?></td>

<td> <?php echo("<a href='delete.php?user_id=" . $d_row["FormId"] . " \"onclick=\"return confirm('Are you sure?')\"'>Delete</a>"); ?></td>

    Thanks

    deriving age from someone's birthday

    in PHP Coding Help

    Posted

    @Barand...….genius! I should have tried that......popped into my head a few days ago but for some reason I never tried to combine them!

    Yes it was from the same book....believe it or not ( I know it doesn't show on these forums) but I got a 2.1 in Software engineering at Uni. Just didn't code for 14 years after that.

     

     

    deriving age from someone's birthday

    in PHP Coding Help

    Posted

    Come on barand, look i'll explain. This is what I have: 

    $query = "SELECT 
 Form.FormId, 
 Form.FirstName, 
 Form.LastName, 
 Form.Email, 
 Form.Age, 
 Form.Birthdate, 
 Form.FavLanguage, 
 GROUP_CONCAT(Vehicle.VehSelection SEPARATOR ', ') AS VehSelection
FROM 
  Vehicle 
  RIGHT JOIN Form 
    ON Form.FormId = Vehicle.FormId
GROUP BY 
  Form.FormId";
  
$result = mysqli_query($link, $query);

$agequery = "SELECT timestampdiff(year, Birthdate, curdate()) as age2 FROM Form";

$resultage = mysqli_query($link, $agequery);

    The first $query puts all the VehSelections on one line. Then I save that into $result. Then I do another query called $agequery and save it into $resultage.  Then I fetch the $result and save it to the $table array. 

      $table = [];
	while ($row = mysqli_fetch_assoc($result)) {
		$table[] = $row;

	}

    So then can I save both results ($result and $resultage) into the same $table array? Because I'm not sure if you can do that. I tried but it didn't work(maybe i'm doing something wrong. So instead I create a new table array, called $tableage: 

    
$tableage = [];
	while ($rowage = mysqli_fetch_assoc($resultage)) {
		$tableage[] = $rowage;

	}

    So moving on from there, If I have $table and $tableage, I got the derived ages in $tableage and the rest of the data in $table. So when I loop round to print the records from the database I use: 

    <?php 

	if ($table) {
		foreach ($table as $d_row) {
			?>
			<tr>
				<td><?php echo($d_row["FormId"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["FirstName"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["LastName"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["Email"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["Age"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["Birthdate"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["FavLanguage"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["VehSelection"]); ?></td>
				<td width="10"> </td>
				<td><?php echo("<a href='edit3.php?user_id=" . $d_row["FormId"] . "'>Edit</a>"); ?></td>
				<td width="10"> </td>
				<td> <?php echo("<a href='delete_feedback.php?user_id=" . $d_row["FormId"] . "'>Delete</a>"); ?></td>
			</tr>
			<?php 
		}
	}

?>

    But because the foreach says $table, I can't use $tableage in there as well. So I start a new foreach: 

    <?php 
	if ($tableage) {
		foreach ($tableage as $d_row1) {
			?>
 <td><strong>new Age</strong></td>
		<td width="10"> </td>
	<td><?php echo($d_row1["age2"]); ?></td>
	<?php 
}}

?>

    My question is how do I combine the printed out results into one? I would like to replace  

    		<td><?php echo($d_row["Age"]); ?></td>

    with  

    <td><?php echo($d_row1["age2"]); ?></td>

    And have the printed results in one foreach, not two.

     

    deriving age from someone's birthday

    in PHP Coding Help

    Posted

    I guess the real question was how do I get the derived ages into my View page? Because I had the SQL and it was working, all the guys who posted here posted code that derived the age correctly but I couldn't get it into my View page which looks like this: 

    	<?php 

	if ($table) {
		foreach ($table as $d_row) {
			?>
			<tr>
				<td><?php echo($d_row["FormId"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["FirstName"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["LastName"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["Email"]); ?></td>
				<td width="10"> </td>
				<td><?php echo $diff->format("%y years"); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["Birthdate"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["FavLanguage"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["VehSelection"]); ?></td>
				<td width="10"> </td>
				<td><?php echo("<a href='edit3.php?user_id=" . $d_row["FormId"] . "'>Edit</a>"); ?></td>
				<td width="10"> </td>
				<td> <?php echo("<a href='delete_feedback.php?user_id=" . $d_row["FormId"] . "'>Delete</a>"); ?></td>
			</tr>
			<?php 
		}
	}
?>

    That was the problem from the start, as before I posted I had the correct sql but couldn't manage to get it into my View page, which displays all records from the database. I was doing this: 

    	<?php 

	if ($table) {
		foreach ($table as $d_row) {
			?>
			<tr>
				<td><?php echo($d_row["FormId"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["FirstName"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["LastName"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["Email"]); ?></td>
				<td width="10"> </td>
						<?php 
		}
	}
?>
<?php
foreach ($resultage as $row) {
	?>
   
  
				<td><?php  echo " {$row['age2']} <br>"; ?></td>
		<?php	} ?>
			<?php
			foreach ($table as $d_row) {
			?>
				<td width="10"> </td>
				<td><?php echo($d_row["Birthdate"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["FavLanguage"]); ?></td>
				<td width="10"> </td>
				<td><?php echo($d_row["VehSelection"]); ?></td>
				<td width="10"> </td>
				<td><?php echo("<a href='edit3.php?user_id=" . $d_row["FormId"] . "'>Edit</a>"); ?></td>
				<td width="10"> </td>
				<td> <?php echo("<a href='delete_feedback.php?user_id=" . $d_row["FormId"] . "'>Delete</a>"); ?></td>
			</tr>
			<?php 
		}
	
?>

    But when I loaded the page it looked a total mess.

    So I'm using Barands code at the moment which is this: 

    $agequery = "SELECT timestampdiff(year, Birthdate, curdate()) as age2 FROM Form";

$resultage = mysqli_query($link, $agequery);



foreach ($resultage as $row) {
    echo " {$row['age2']} <br>";
}

    Now how do I get echo " {$row['age2']} <br>";  into the top block of code in this post, replacing     <td><?php echo $diff->format("%y years"); ?></td>      ?

    deriving age from someone's birthday

    in PHP Coding Help

    Posted

    1 minute ago, ginerjm said:

    Yes all of the formids were probably stored in there but you wrote a query with a WHERE clause asking for one record that matched whatever was in the variable named $FormId.

    I just thought it would take each FormId, 1 by 1, and calculate the age for each record and print it out

    I understand what your saying about variables.

    deriving age from someone's birthday

    in PHP Coding Help

    Posted

    Just now, Barand said:

    When you do a query to select only 1 record, why would you expect output for more than one record?

    I didn't know it was only querying one record, I thought all the FormId's from the database were stored in there,  as I've done similar sql code like that before and it worked

    deriving age from someone's birthday

    in PHP Coding Help

    Posted

    Thanks to both of your replies. That's what I did Barand, I used: 

    SELECT TIMESTAMPDIFF (YEAR, Birthdate, CURDATE()) AS AGE FROM Form;

    It gave me correct results for all the records, but when I tried to put it into my code I did this: 

    SELECT TIMESTAMPDIFF (YEAR, Birthdate, CURDATE()) AS Age1 FROM Form WHERE Form.FormId = '$FormId'";

    But when I tried to loop round it and output the records I only ever got result for first record. I'll try out solutions from above. Thanks

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