php - validate if table row has data in every fields
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I want to validate if a current row in every field has data on it and if it has not then display no data. Please help me. Also it is advisable to use echo in displaying the table data or should i stick to the html. Thanks in advance! I've been searching the net and was not able to find some answers. lol
$ct_list = mysqli_query($conn,"select * from ct LEFT JOIN station on ct.station_id=station.station_id LEFT JOIN dilg_emp on dilg_emp.station_id=station.station_id where dilg_emp.dilg_emp_id= $session_id");
if (mysqli_num_rows($ct_list) > 0){
while ($ct_row_list = mysqli_fetch_array($ct_list)) {
$ct_FName = $ct_row_list['ct_FName'];
$ct_MName = $ct_row_list['ct_MName'];
$ct_LName = $ct_row_list['ct_LName'];
$ct_SName = $ct_row_list['ct_SName'];
$ct_address = $ct_row_list['ct_address'];
$ct_birthday = $ct_row_list['ct_birthday'];
$ct_age = $ct_row_list['ct_age'];
$ct_gender = $ct_row_list['ct_gender'];
$ct_level_educ = $ct_row_list['ct_level_educ'];
$ct_course = $ct_row_list['ct_course'];
$ct_school = $ct_row_list['ct_school'];
$ct_prev_covid = $ct_row_list['ct_prev_covid'];
$ct_training_date = $ct_row_list['ct_training_date'];
$ct_deployment_date = $ct_row_list['ct_deployment_date'];
$ct_prev_emp = $ct_row_list['ct_prev_emp'];
echo "<tr>
<td>".
$x++."
</td>
<td>
$ct_FName $ct_MName $ct_LName $ct_SName
</td>
<td>
$ct_address
</td>
<td>
$ct_birthday
</td>
<td>
$ct_age
</td>
<td>
$ct_gender
</td>
<td>
$ct_level_educ
</td>
<td>
$ct_course
</td>
<td>
$ct_school
</td>
<td>
$ct_prev_covid
</td>
<td>
$ct_training_date
</td>
<td>
$ct_deployment_date
</td>
<td>
$ct_prev_emp
</td>
</tr>";
}
}else{
echo "<tr><td><b>Nothing to display</b></td></tr>";
}
Answer
Solution:
You can loop over the row array, replacing empty values withNo data.
$optional_fields = ["ct_SName"];
while ($ct_row_list = mysqli_fetch_array($ct_list)) {
foreach ($ct_row_list as $name => &$value) {
if (in_array($name, $optional_fields)) {
continue;
} elseif (empty($value)) {
$value = "No data";
}
}
// rest of your code goes here
Answer
Solution:
It's hard to tell exactly what you mean, if a row has any empty columns then skip that row?
If you select 15 columns in the query or if you know thatSELECT * will return 15 columns, then filter out the empty ones and see if there are 15:
if (mysqli_num_rows($ct_list) > 0) {
while ($ct_row_list = mysqli_fetch_array($ct_list)) {
if (count(array_filter($ct_row)) == 15) {
echo "<tr>
<td>$x++</td>
<td>$ct_FName $ct_MName $ct_LName $ct_SName</td>"; //etc...
}
}
}
You could also check and skip:
if (count(array_filter($ct_row)) < 15) {
continue;
}
//echo your HTML
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