需要连接TWO表PHP / MYSQLI(Need connection between TWO tables PHP/MYSQLI)

这是我的代码：

<?php $target_dir = "videoCover/"; $target_dir = $target_dir . basename( $_FILES["uploadFile"]["name"]); $uploadOk=1; // Check if file already exists if (file_exists($target_dir . $_FILES["uploadFile"]["name"])) { echo "Sorry, file already exists."; $uploadOk = 0; } // Check file size if ($uploadFile_size > 128000000) { echo "Sorry, your file is too large."; $uploadOk = 0; } // Check if $uploadOk is set to 0 by an error if ($uploadOk==0) { echo "Sorry, your file was not uploaded."; // if everything is ok, try to upload file } else { if (move_uploaded_file($_FILES["uploadFile"]["tmp_name"], $target_dir)) { echo "The file ". basename( $_FILES["uploadFile"]["name"]). " has been uploaded."; } else { echo "Sorry, there was an error uploading your file."; } } include '../connect/con.php'; $id = mysqli_real_escape_string($con, $_POST['id']); $vidTitle = mysqli_real_escape_string($con, $_POST['vidTitle']); $imgCover = $_FILES['uploadFile']['name']; $size = $_FILES['uploadFile']['size']; $type = $_FILES['uploadFile']['type']; $url = '/upload/videoCover/'.$_FILES["uploadFile"]["name"]; $vidSD = mysqli_real_escape_string($con, $_POST['vidSD']); $sql="INSERT INTO newsvid (id, vidTitle, imgCover, size, type, url, vidSD) VALUES ('$id', '$vidTitle', '$imgCover', '$size', '$type', '$url', '$vidSD')"; if (!mysqli_query($con,$sql)) { die('Error: ' . mysqli_error($con)); }echo "Video links are added"; mysqli_close($con); ?> <?php include '../connect/con.php'; if(isset($_FILES['file_array'])){ $name_array = $_FILES['file_array']['name']; $tmp_name_array = $_FILES['file_array']['tmp_name']; $type_array = $_FILES['file_array']['type']; $size_array = $_FILES['file_array']['size']; $error_array = $_FILES['file_array']['error']; for($i = 0; $i < count($tmp_name_array); $i++){ if(move_uploaded_file($tmp_name_array[$i], "videoScreenShots/".$name_array[$i])){ echo $name_array[$i]." upload is complete<br>"; $imgShot[$i]='videoScreenShots/'.$name_array[$i]; } else { echo "move_uploaded_file function failed for ".$name_array[$i]."<br>"; } } } $idvi = mysqli_insert_id($con); $vidLD = mysqli_real_escape_string($con, $_POST['vidLD']); $vidYear = mysqli_real_escape_string($con, $_POST['vidYear']); $vidCity = mysqli_real_escape_string($con, $_POST['vidCity']); $vidZanr = mysqli_real_escape_string($con, $_POST['vidZanr']); $vidZanr2 = mysqli_real_escape_string($con, $_POST['vidZanr2']); $vidZanr3 = mysqli_real_escape_string($con, $_POST['vidZanr3']); $vidQuality = mysqli_real_escape_string($con, $_POST['vidQuality']); $vidTranslated = mysqli_real_escape_string($con, $_POST['vidTranslated']); $vidTime = mysqli_real_escape_string($con, $_POST['vidTime']); $vidMaker = mysqli_real_escape_string($con, $_POST['vidMaker']); $vidRoles = mysqli_real_escape_string($con, $_POST['vidRoles']); $sql="INSERT INTO videoinformation (id, vidLD, vidYear, vidCity, vidZanr, vidZanr2, vidZanr3, vidQuality, vidTranslated, vidTime, vidMaker, vidRoles, imgShot1, imgShot2, imgShot3) VALUES ('$idvi', '$vidLD', '$vidYear', '$vidCity', '$vidZanr', '$vidZanr2', '$vidZanr3', '$vidQuality', '$vidTranslated', '$vidTime', '$vidMaker', '$vidRoles', '".$imgShot[0]."','".$imgShot[1]."','".$imgShot[2]."')"; if (!mysqli_query($con,$sql)) { die('Error: ' . mysqli_error($con)); }echo "Video Description are added"; mysqli_close($con); ?>

问题是我正在使用这个$idvi = mysql_insert_id($con); 从首次上传信息到第一个表（newsvid）获取ID。

我需要的是：我的表单有很多不同的信息，我尝试将所有这些信息上传到两个不同的表中。在第一次上传到table1并使用table1中的id上传信息到table2的方式。最后我被困住了，不知道怎么做（

PS目前使用此代码我只有空/空白页面。

PPS有错误： Error: Cannot add or update a child row: a foreign key constraint fails (denzw681_u.videoinformation, CONSTRAINT videoinformation_ibfk_1 FOREIGN KEY (id) REFERENCES newsvid (id))

This is my code:

<?php $target_dir = "videoCover/"; $target_dir = $target_dir . basename( $_FILES["uploadFile"]["name"]); $uploadOk=1; // Check if file already exists if (file_exists($target_dir . $_FILES["uploadFile"]["name"])) { echo "Sorry, file already exists."; $uploadOk = 0; } // Check file size if ($uploadFile_size > 128000000) { echo "Sorry, your file is too large."; $uploadOk = 0; } // Check if $uploadOk is set to 0 by an error if ($uploadOk==0) { echo "Sorry, your file was not uploaded."; // if everything is ok, try to upload file } else { if (move_uploaded_file($_FILES["uploadFile"]["tmp_name"], $target_dir)) { echo "The file ". basename( $_FILES["uploadFile"]["name"]). " has been uploaded."; } else { echo "Sorry, there was an error uploading your file."; } } include '../connect/con.php'; $id = mysqli_real_escape_string($con, $_POST['id']); $vidTitle = mysqli_real_escape_string($con, $_POST['vidTitle']); $imgCover = $_FILES['uploadFile']['name']; $size = $_FILES['uploadFile']['size']; $type = $_FILES['uploadFile']['type']; $url = '/upload/videoCover/'.$_FILES["uploadFile"]["name"]; $vidSD = mysqli_real_escape_string($con, $_POST['vidSD']); $sql="INSERT INTO newsvid (id, vidTitle, imgCover, size, type, url, vidSD) VALUES ('$id', '$vidTitle', '$imgCover', '$size', '$type', '$url', '$vidSD')"; if (!mysqli_query($con,$sql)) { die('Error: ' . mysqli_error($con)); }echo "Video links are added"; mysqli_close($con); ?> <?php include '../connect/con.php'; if(isset($_FILES['file_array'])){ $name_array = $_FILES['file_array']['name']; $tmp_name_array = $_FILES['file_array']['tmp_name']; $type_array = $_FILES['file_array']['type']; $size_array = $_FILES['file_array']['size']; $error_array = $_FILES['file_array']['error']; for($i = 0; $i < count($tmp_name_array); $i++){ if(move_uploaded_file($tmp_name_array[$i], "videoScreenShots/".$name_array[$i])){ echo $name_array[$i]." upload is complete<br>"; $imgShot[$i]='videoScreenShots/'.$name_array[$i]; } else { echo "move_uploaded_file function failed for ".$name_array[$i]."<br>"; } } } $idvi = mysqli_insert_id($con); $vidLD = mysqli_real_escape_string($con, $_POST['vidLD']); $vidYear = mysqli_real_escape_string($con, $_POST['vidYear']); $vidCity = mysqli_real_escape_string($con, $_POST['vidCity']); $vidZanr = mysqli_real_escape_string($con, $_POST['vidZanr']); $vidZanr2 = mysqli_real_escape_string($con, $_POST['vidZanr2']); $vidZanr3 = mysqli_real_escape_string($con, $_POST['vidZanr3']); $vidQuality = mysqli_real_escape_string($con, $_POST['vidQuality']); $vidTranslated = mysqli_real_escape_string($con, $_POST['vidTranslated']); $vidTime = mysqli_real_escape_string($con, $_POST['vidTime']); $vidMaker = mysqli_real_escape_string($con, $_POST['vidMaker']); $vidRoles = mysqli_real_escape_string($con, $_POST['vidRoles']); $sql="INSERT INTO videoinformation (id, vidLD, vidYear, vidCity, vidZanr, vidZanr2, vidZanr3, vidQuality, vidTranslated, vidTime, vidMaker, vidRoles, imgShot1, imgShot2, imgShot3) VALUES ('$idvi', '$vidLD', '$vidYear', '$vidCity', '$vidZanr', '$vidZanr2', '$vidZanr3', '$vidQuality', '$vidTranslated', '$vidTime', '$vidMaker', '$vidRoles', '".$imgShot[0]."','".$imgShot[1]."','".$imgShot[2]."')"; if (!mysqli_query($con,$sql)) { die('Error: ' . mysqli_error($con)); }echo "Video Description are added"; mysqli_close($con); ?>

Question is i'm using this $idvi = mysql_insert_id($con); to get ID from first uploading information to first table (newsvid).

All what I need is: I have form which have a lot of different information and I try to upload all this information into two different tables. In the way like uploading first to table1 and using id from table1 upload information to table2. Finally I'm stuck and do not get how I can do it (

P.S. Currently using this code I have just empty/blank page.

P.P.S Have an error: Error: Cannot add or update a child row: a foreign key constraint fails (denzw681_u.videoinformation, CONSTRAINT videoinformation_ibfk_1 FOREIGN KEY (id) REFERENCES newsvid (id))