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Hello This secret code will not be published into the live site, just a test copy The below code worked with me if(isset($_POST['submitContact']) && $_SERVER['REQUEST_METHOD'] == 'POST'){ $postdata =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>"; }

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 ) )

It was inside the () just I wrote here by wrong

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>"; }

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

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>"; }

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

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

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);

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 ?>

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

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

Yes, I will do this, should be better, thanks

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

You mean that serialized, base64encoded is not a good practice for such storing ?