How do I display a single entry from an array from MYSQL

This works ...

$result = mysqli_query($conn, "SELECT * from $table");
          $pix = array();
          while ($row = $result->fetch_assoc()) {
           $pix[] = $row;
          }
          foreach ($pix as $pic) {
           echo "PID: {$pic['PID']}<br>"
            . "Title: {$pic['title']}<br>"
            . "Filename: {$pic['filename']}<br><br>";
          }

$result = mysqli_query($conn, "SELECT * from $table");
          $pix = array();
          while ($row = $result->fetch_assoc()) {
           $pix[] = $row;
          }
          foreach ($pix as $pic) {
           echo "PID: {$pic['PID']}<br>"
            . "Title: {$pic['title']}<br>"
            . "Filename: {$pic['filename']}<br><br>";
          }

How do I display just the entry with PID value of "5" from the array?

19 Replies

Jochem•3y ago

SELECT * FROM table_name WHERE PID = 5;

SELECT * FROM table_name WHERE PID = 5;

alternatively if you really have to do it in PHP:

foreach ($pix as $pic) {
  if ($pic['PID'] != 5) continue;

  echo "PID: {$pic['PID']}<br>"
  . "Title: {$pic['title']}<br>"
  . "Filename: {$pic['filename']}<br><br>";
}

foreach ($pix as $pic) {
  if ($pic['PID'] != 5) continue;

  echo "PID: {$pic['PID']}<br>"
  . "Title: {$pic['title']}<br>"
  . "Filename: {$pic['filename']}<br><br>";
}

BlackwolfOP•3y ago

I need the whole database in the array I am then dynamically changing data on the screen

Jochem•3y ago

the if-statement I added would leave the array alone, but only print out info if $pic['PID'] is 5

BlackwolfOP•3y ago

oh right, i thought there would be an easier way like echo $pix[5, 'PID']; 5 being the 5th entry in the array

Jochem•3y ago

you can't guarantee the one with PID 5 will always be the fifth one though

BlackwolfOP•3y ago

so $pix[5,'title'] would be the title of the 5th entry

Jochem•3y ago

if all you want is the ~~fifth~~sixth element, you can just do $pix[5]['PID'] though

BlackwolfOP•3y ago

ahhhhh cool, it was getting the format correct 🙂

Jochem•3y ago

and remember arrays are 0-indexed, so 0 is the first element just remember that there's no direct relation to a field in the database and the index of the array. If PID 2 is ever missing, all of a sudden you're getting the wrong element if you really need PID 5

BlackwolfOP•3y ago

oh yea, but PID was an example of being specific i am listing titles in a <select> and when an option changes I want to display the elements for that opion option* when an option is selected, not changed

Jochem•3y ago

well, the syntax is like I said if you want to access elements inside an array, and you can chain the square brackets to go deeper if you have deeper arrays 🙂

BlackwolfOP•3y ago

Thank you kindly sir, that works a treat 🙂

Jochem•3y ago

no worries 🙂

BlackwolfOP•3y ago

if I know the title, can I display the corresponding data? so if I have a picture called "Scream" ... would $pix['Scream']['filename'] work

Jochem•3y ago

oh wait, you think $pix[5]['PID'] filters the array and finds a record that has 5 as value for the PID? what it's doing is accessing the $pix array, taking the sixth (starting from 0, 5 is the sixth element) element, and returning that. ['PID'] then gives you the value for that element's PID key so no, $pix['Scream']['filename'] doesn't filter for elements with filename scream, it would try to take the 'Scream' element from the pix array, and then the filename element from that Scream array

BlackwolfOP•3y ago

i understand that $pix[5]['PID'] is taking the 6 line of the array i was just wondering if i could use actualy data to find other data actual* so to find the date of when the picture "Scream" was posted for example

Jochem•3y ago

the result from fetch_assoc would contain indexed rows, and the rows themselves would have named keys, so no, $pix['Scream'] wouldn't work without some manipulation of the result from mysql

BlackwolfOP•3y ago

ok, i understand, thank you so my best bet is to simply put the array value in the <option> and use that 😉 so this works beautifully, thank you so much 🙂

// load whole db into array
          $result = mysqli_query($conn, "SELECT * from $table ORDER BY title");
          $pix = array();
          while ($row = $result->fetch_assoc()) {
           $pix[] = $row;
          }
          // echo $pix[5]['title'];


          // create <select> options
          $result = mysqli_query($conn, "SELECT * from $table ORDER BY title");
          if ($result->num_rows > 0) {
           $count = "0";
           while ($row = $result->fetch_assoc()) {
            echo '
            <option value="' . $count . '">' . $row['title'] . '</option>';
            $count++;
           }
          }
          $conn->close();
          ?>
         </select>
         <?php
         echo $pix[5]['title'];
         ?>

// load whole db into array
          $result = mysqli_query($conn, "SELECT * from $table ORDER BY title");
          $pix = array();
          while ($row = $result->fetch_assoc()) {
           $pix[] = $row;
          }
          // echo $pix[5]['title'];


          // create <select> options
          $result = mysqli_query($conn, "SELECT * from $table ORDER BY title");
          if ($result->num_rows > 0) {
           $count = "0";
           while ($row = $result->fetch_assoc()) {
            echo '
            <option value="' . $count . '">' . $row['title'] . '</option>';
            $count++;
           }
          }
          $conn->close();
          ?>
         </select>
         <?php
         echo $pix[5]['title'];
         ?>

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How do I display a single entry from an array from MYSQL