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amirelgohary1990

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Everything posted by amirelgohary1990

  1. amirelgohary1990
    I am retrieving data via Ajax into Choices.js select options My scenario is when I select date, getting the available restaurant tables The retrieving data is 100% working, but it reflects in the select options only in the first request, then when I try to reselect another date I receive the below console error, and choices still keep the initial retrieved data choices.min.js:11 Trying to initialise Choices on element already initialised Choices setValue Function var el = document.getElementsByClassName("table_number")[0]; if (el) { function setChoices(values) { const tableNumbers = new Choices(el, { removeItemButton: true, }).setValue(values); } setChoices(values); } Ajax Code let shiftDate = document.getElementById('reservation_shift_date'); shiftDate.addEventListener("change", function(){ let request = new XMLHttpRequest(); request.open("POST","get_tables.php",true); request.setRequestHeader("content-type","application/x-www-form-urlencoded"); request.onreadystatechange = function(){ if(request.readyState == 4 && request.status == 200){ setChoices(JSON.parse(request.responseText)); } } request.send("date="+shiftDate.value); }); get_tables.php if($_SERVER["REQUEST_METHOD"] === "POST" && isset($_POST['date'])){ $stmt = $conn->prepare("SELECT table_id FROM reservation WHERE shift_date = ?"); $stmt->execute([$_POST['date']]); $rows = $stmt->fetchAll(PDO::FETCH_ASSOC); foreach($rows as $row){ $reserved_tables_id[] = $row['table_id']; } $in = implode(',',$reserved_tables_id); $execute_query = mysqli_query($dbConnection,"SELECT id, table_name FROM tables WHERE id NOT IN ($in)"); while($row = mysqli_fetch_assoc($execute_query)){ $tbl_id = $row['id']; $tbl_name = $row['table_name']; $arr[] = ["value"=>$tbl_id,"label"=>$tbl_name]; } echo json_encode($arr); }
  2. amirelgohary1990
    Hello This secret code will not be published into the live site, just a test copy The below code worked with me if(isset($_POST['submitContact']) && $_SERVER['REQUEST_METHOD'] == 'POST'){ $postdata = http_build_query(["secret"=>"6LfyPF0pAAAAAEsS5lfN_WL3wKHh1XfGo0oE_PYU","response"=>$recaptcha_response]); $opts = ['http' => [ 'method' => 'POST', 'header' => 'Content-type: application/x-www-form-urlencoded', 'content' => $postdata ] ]; $context = stream_context_create($opts); $result = file_get_contents('https://www.google.com/recaptcha/api/siteverify', false, $context); $recaptcha = json_decode($result); if($recaptcha->success ==true){ if($recaptcha->score >= 0.5){ echo "Recaptcha Success"; }else{ echo"<pre>"; print_r("Recaptcha Not Verified"); echo"</pre>"; } }else{ echo"<pre>"; print_r($recaptcha); echo"</pre>"; }
  3. amirelgohary1990
    Hello I am receiving a huge amount of spam emails, now I am trying to implement Google Recaptcha V3 in my custom PHP From, I implemented all the steps for G-Recaptcha, but I receive error invalid-input-secret And I am sure that the secret code shout be copied right I added the below to the head tag <script src="https://www.google.com/recaptcha/api.js?render=6LfyPF0pAAAAAHLxp3315RTN7jrRvBe6kLdHGAiT"></script> <script> grecaptcha.ready(function() { grecaptcha.execute('6LfyPF0pAAAAAHLxp3315RTN7jrRvBe6kLdHGAiT', {action: 'submit'}).then(function(token) { let recaptchaResponse = document.getElementById("recaptchaResponse"); console.log(recaptchaResponse); recaptchaResponse.value = token; }); }); </script> Then added hidden input before the submit button in the Form <input type="hidden" name="recaptcha_response" id="recaptchaResponse"> <input class="contactInput no-border cursorPointer buttonStyle" name="submitContact" value="Submit" type="submit"> And finally, I implemented the PHP code if(isset($_POST['submitContact']) && $_SERVER['REQUEST_METHOD'] == 'POST'){ $recaptcha_url = 'https://www.google.com/recaptcha/api/siteverify'; $recaptcha_secret = '6LfyPF0pAAAAAEsS5lfN_WL3wKHh1XfGo0oE_PYU'; $recaptcha_response = $_POST['recaptcha_response']; $recaptcha = file_get_contents($recaptcha_url."?secret=".$recaptcha_secret."?response=".$recaptcha_response); $recaptcha = json_decode($recaptcha); if($recaptcha->success ==true){ if($recaptcha->score >= 0.5){ echo "Recaptcha Success"; }else{ echo"<pre>"; print_r("Recaptcha Not Verified"); echo"</pre>"; } }else{ echo"<pre>"; print_r($recaptcha); echo"</pre>"; } } But receiving the below error stdClass Object ( [success] => [error-codes] => Array ( [0] => invalid-input-secret ) )
  4. amirelgohary1990
    It was inside the () just I wrote here by wrong
  5. amirelgohary1990
    I tried this also, but same problem $bloodType_list = ['a+','a-','b+','b-','o+','o-','ab+','ab-']; if($key = array_search('a+',$bloodType_list)){ if($key !== false){ unset($bloodType_list[$key]); } } foreach($bloodType_list as $bloodType_lists){ echo $bloodType_lists."<br>"; }
  6. amirelgohary1990
    You mean like below code ? $bloodType_list = ['a+','a-','b+','b-','o+','o-','ab+','ab-']; if($key = array_search('a+',$bloodType_list)) !== false { unset($bloodType_list[$key]); } foreach($bloodType_list as $bloodType_lists){ echo $bloodType_lists."<br>"; } I tried that not working too
  7. amirelgohary1990
    Hello, I am trying to unset from array using array_search, it's working, except the first array value "a+" is not working $bloodType_list = ['a+','a-','b+','b-','o+','o-','ab+','ab-']; if($key = array_search('a+',$bloodType_list)){ unset($bloodType_list[$key]); } foreach($bloodType_list as $bloodType_lists){ echo $bloodType_lists."<br>"; }
  8. amirelgohary1990
    Hello, photos do not appear in the following cases 1- inserting photo inside <picture> tag (In Server Side Only), Working normally in local 2- Not Working when site domain only www.mysite.com ,, Working when site domain = www.mysite.com/index.php <picture class="hover1 runHover2"> <img src="assets/img/central-business-district-singapore.jpg"> </picture> Note: -The path is right, and when opening inspect to check the photo, I find it - Tried to change the path to /assets/img/central-business-district-singapore.jpg or www.mysite.com/assets/img/central-business-district-singapore.jpg , but not solved
  9. amirelgohary1990
    1- Noted, and edited 2- I think this is the last point I need, I added the imploded ids inside mysqli_stmt_bind_param, is this right, or I have to put somewhere else,I tried even to put my integers manually inside the FIND_IN_SET('id',77,181), but did not work gives me error Warning: mysqli_stmt_bind_param() expects parameter 1 to be mysqli_stmt, bool given 3- Actually $the_array = [77,181]; I wrote here as static, but in my project this array comes from another dynamic query, that comes from the website when user search for specific country, Regarding
  10. amirelgohary1990
    Code not returning anything Yes, I enabled errors, also no errors, but I am sure something wrong, when I switch to normal query everything works normally with IN() ini_set('display_errors', 1); ini_set('display_startup_errors', 1); error_reporting(E_ALL);
  11. amirelgohary1990
    I tried to use FIND_IN_SET() with prepared statement, but did work, do not return any result, or even errors if(escape($_POST['jobCategory']) != "all-categories" && escape($_POST['countryId']) == "all-countries" && escape($_POST['careerLevel']) == "all-career-levels"): $the_array = [77,181]; $job_id_imploded = implode(',',$the_array); $query = mysqli_prepare($dbConnection,"SELECT jobs.id, jobs.job_title, jobs.country_id, employers.employer_name FROM jobs LEFT JOIN employers ON jobs.employer_id = employers.employer_id WHERE job_status = ? AND FIND_IN_SET('id',?)"); mysqli_stmt_bind_param($query,'si',$job_status,$job_id_imploded); endif; mysqli_stmt_execute($query); mysqli_stmt_bind_result($query,$job_id,$job_title,$countryId,$employer_name); while(mysqli_stmt_fetch($query)){ ?> <div class="job-title"> <a href="job_post.php?job_id=<?php echo htmlspecialchars($job_id) ?>" class="job-title-link"><?php echo htmlspecialchars($job_title); ?></a> </div> <?php } // End While ?>
  12. amirelgohary1990
    I tried to use FIND_IN_SET() with prepared statement, but did work, do not return any result, or even errors, that's why I used normal query until study PDO then switch this query into PDO, I will open new case with my code with FIND_IN_SET() may be something wrong in my code
  13. amirelgohary1990
    Hello I need to find a way to close loop outside if condition like below example if(escape($_POST['jobCategory']) != "all-categories" && escape($_POST['countryId']) == "all-countries"): $query = mysqli_query($dbConnection,"SELECT jobs.id, jobs.job_title, jobs.salary, jobs.employer_id, employers.employer_name, employers.employer_logo FROM jobs LEFT JOIN employers ON jobs.employer_id = employers.employer_id WHERE job_status = '".mysqli_real_escape_string($dbConnection,'Active')."' AND id IN (".mysqli_real_escape_string($dbConnection,$job_id_imploded).") "); while($row = mysqli_fetch_assoc($query)){ // Start Loop $job_id = $row['id']; $job_title = $row['job_title']; endif; <div class="job-title"> <a href="job_post.php?job_id=<?php echo htmlspecialchars($job_id) ?>" class="job-title-link"><?php echo htmlspecialchars($job_title); ?></a> </div> } // End Of Loop Gives me error HTTP ERROR 500
  14. amirelgohary1990
    Yes, I will do this, should be better, thanks
  15. amirelgohary1990
    I started using serialize because I wanted to store multiple steps of values to create dynamic taxes for a HRM system I searched for a way I found that serialize can be used if these data will not be used for analytics, just will sore it, then later I retrieve data as array normally
  16. amirelgohary1990
    You mean that serialized, base64encoded is not a good practice for such storing ?
  17. amirelgohary1990
    Hello, I am selecting two rows, first is 'id' and second row I am selecting serialized 'category_ids', then I unserialized it Those two rows I selected 'id' + unserialized 'category_ids' ,, I am trying to output then into specific array template like the below array example The Array "I Need to achieve like this" ["172"=>["4","6"],"174"=>["4","6"],"175"=>["4","3","6"],"176"=>["4","3"],"177"=>["4","6"],"181"=>["3","6"],"182"=>["7"],"183"=>["3","4"],"184"=>["4","3","6"],"185"=>["3","6"],"186"=>["8","6"],"188"=>["3","6"],"189"=>["3","6"],"190"=>["6"],"191"=>["3","6","4","7"]]; It's var_dump "I Need to achieve like this" array(15 { [172]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "6"} [174]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "6" } [175]=> array(3) { [0]=> string(1) "4" [1]=> string(1) "3" [2]=> string(1) "6" } [176]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "3" } [177]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "6" } [181]=> array(2) { [0]=> string(1) "3" [1]=> string(1) "6" } [182]=> array(1) { [0]=> string(1) "7" } [183]=> array(2) { [0]=> string(1) "3" [1]=> string(1) "4" } [184]=> array(3) { [0]=> string(1) "4" [1]=> string(1) "3" [2]=> string(1) "6" } [185]=> array(2) { [0]=> string(1) "3" [1]=> string(1) "6" } [186]=> array(2) { [0]=> string(1) "8" [1]=> string(1) "6" } [188]=> array(2) { [0]=> string(1) "3" [1]=> string(1) "6" } [189]=> array(2) { [0]=> string(1) "3" [1]=> string(1) "6" } [190]=> array(1) { [0]=> string(1) "6" } [191]=> array(4) { [0]=> string(1) "3" [1]=> string(1) "6" [2]=> string(1) "4" [3]=> string(1) "7" } } When I try to achieve this by the following $asscArrays = [$job_id=>$job_category_id]; It's printing the array result but separately array(1) not array(15) ,, like the below array(1) { [172]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "6" } } array(1) { [174]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "6" } } array(1) { [175]=> array(3) { [0]=> string(1) "4" [1]=> string(1) "3" [2]=> string(1) "6" } } array(1) { [176]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "3" } } array(1) { [177]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "6" } } array(1) { [181]=> array(2) { [0]=> string(1) "3" [1]=> string(1) "6" } } array(1) { [182]=> array(1) { [0]=> string(1) "7" } } The array template I am willing to achieve is array(15)
  18. amirelgohary1990
    I need to output the exact below array format from the mysqli_fetch_assoc rows The Array "Need to achieve like this" ["172"=>["4","6"],"174"=>["4","6"],"175"=>["4","3","6"],"176"=>["4","3"],"177"=>["4","6"],"181"=>["3","6"],"182"=>["7"],"183"=>["3","4"],"184"=>["4","3","6"],"185"=>["3","6"],"186"=>["8","6"],"188"=>["3","6"],"189"=>["3","6"],"190"=>["6"],"191"=>["3","6","4","7"]]; It's var_dump "Need to achieve like this" array(15) { [172]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "6"} [174]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "6" } [175]=> array(3) { [0]=> string(1) "4" [1]=> string(1) "3" [2]=> string(1) "6" } [176]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "3" } [177]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "6" } [181]=> array(2) { [0]=> string(1) "3" [1]=> string(1) "6" } [182]=> array(1) { [0]=> string(1) "7" } [183]=> array(2) { [0]=> string(1) "3" [1]=> string(1) "4" } [184]=> array(3) { [0]=> string(1) "4" [1]=> string(1) "3" [2]=> string(1) "6" } [185]=> array(2) { [0]=> string(1) "3" [1]=> string(1) "6" } [186]=> array(2) { [0]=> string(1) "8" [1]=> string(1) "6" } [188]=> array(2) { [0]=> string(1) "3" [1]=> string(1) "6" } [189]=> array(2) { [0]=> string(1) "3" [1]=> string(1) "6" } [190]=> array(1) { [0]=> string(1) "6" } [191]=> array(4) { [0]=> string(1) "3" [1]=> string(1) "6" [2]=> string(1) "4" [3]=> string(1) "7" } } My Code $query = mysqli_query($dbConnection,"SELECT id, job_category_id FROM jobs"); while($row = mysqli_fetch_assoc($query)){ $job_id = $row['id']; $job_category_id = htmlspecialchars($row['job_category_id']); $job_category_id = unserialize(base64_decode($job_category_id)); $asscArrays = [$job_id=>$job_category_id]; // Here I am trying to achieve the array template like the I mentioned above } var_dump $asscArrays array(1) { [172]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "6" } } array(1) { [174]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "6" } } array(1) { [175]=> array(3) { [0]=> string(1) "4" [1]=> string(1) "3" [2]=> string(1) "6" } } array(1) { [176]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "3" } } array(1) { [177]=> array(2) { [0]=> string(1) "4" [1]=> string(1) "6" } } array(1) { [181]=> array(2) { [0]=> string(1) "3" [1]=> string(1) "6" } } array(1) { [182]=> array(1) { [0]=> string(1) "7" } }
  19. amirelgohary1990
    I need to store PDFs in a folders, not keep random in the same index root, Actually it's very strange either path and file 100% are right I double checked
  20. amirelgohary1990
    Hello, When I type admin/assets/cvs/myimage.png It's working but when I type a pdf admin/assets/cvs/myfile.pdf not working That's why I assumed that the path should be right PDF only reads with me if the file in the same folder index like this <a href="myfile.pdf" target="_blank">View PDF</a> This working
  21. amirelgohary1990
    When I try to display PDF , not working when pdf contains path, just working if the file is in the same index This Works <a href="myfile.pdf" target="_blank">View PDF</a> The below not working .. ERROR The requested URL was not found on this server. <a href="admin/assets/cvs/myfile.pdf" target="_blank">View PDF</a> I am sure that the path is right and I tested with .png extensions. Also I tried to use header function to display PDF with PHP but got also error even if the file in the same directory index Failed to load PDF document. $fileName = "myfile.pdf"; header('Content-type: application/pdf'); header('Content-Disposition: inline; filename="' .urlencode($fileName). '"'); header('Content-Transfer-Encoding: binary'); header('Content-Length: ' . filesize($fileName)); header('Accept-Ranges: bytes'); @readfile($fileName);
  22. amirelgohary1990
    @Barand I am new in oop I assumed that $stmt->execute($arr_params); will echo
  23. amirelgohary1990
    Hello @mac_gyver I tried this example in the link you shared Example #5 Execute a prepared statement using array for IN clause Actually the error gone, but did not echo the result from the database $arr_params = [1,2,5]; $placeholders = implode(',', array_fill(0, count($arr_params), '?')); $query = "SELECT designation_id, designation_name FROM designations WHERE designation_id NOT IN ($placeholders)"; try { $stmt = $conn->prepare($query); $stmt->execute($arr_params); $a_data = $stmt->fetchAll(PDO::FETCH_ASSOC); } catch(PDOException $e) { trigger_error('Wrong SQL: ' . $query . ' Error: ' . $e->getMessage(), E_USER_ERROR); }
  24. amirelgohary1990
    Hello, I am trying to use array in bind statement to avoid entering bind manually Below, I set up the array, then imploded the array to insert , on it // to return 1,2,5 , but I got error Warning: mysqli_stmt::bind_param(): Number of elements in type definition string doesn't match number of bind variables $arr = [1,2,5]; $arr_as_string = implode( ',',$arr); $type = 'iii'; $params = [$type,$arr_as_string]; $tmp = []; foreach($params as $key => $value) $tmp[$key] = &$params[$key]; call_user_func_array([$query, 'bind_param'], $tmp);
  25. amirelgohary1990
    Hello Just I am trying to replace $params = array('ss', 'stan', 'Stanley'); With $arr = [1,2,5]; $params = array('ss', 'stan', $arr); But when I do that gives me error Warning: mysqli_stmt::bind_param(): Number of elements in type definition string doesn't match number of bind variables
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