mysqli

问题 I have built a custom CMS for updating my website. Normally what I do is, I mention database login info (like username-host-password) to a php file and include that file into all the php script files having database syntax. By doing this I access database through that specific database user. What I am trying to do is to replace that particular database user with the logged in user. More clearly if user John is logged in to the CMS, he will access the database using the username john. And if

问题 I tried to do multiple updates with mysqli_multi_query() , but it doesn’t work. $arr_updates[] = "UPDATE MYTABLE SET COLUMN = 'VALUE' WHERE ID = 1"; $arr_updates[] = "UPDATE MYTABLE SET COLUMN = 'VALUE' WHERE ID = 2"; $b = mysqli_multi_query($CONN,join("; ",$arr_updates)); $b is FALSE, even though the syntax of the SQL statements is correct. Is mysqli_multi_query() supposed to work with MySQL UPDATE statements? Or is there any way I could optimize my update statement? (I can’t use a switch

问题 I am experiencing issues connecting to a Cloud SQL instance hosted in an App engine (Java Runtime). The connecting client is a PHP web application hosted in another Google app engine instance. Is it possible for an App engine instance to connect to a Cloud SQL in another instance? Basically, my setup is as follow: Google App Engine 1 (Java Runtime) installed with Cloud SQL instance hosts Google Endpoints and connects to Cloud SQL on this instance Google App Engine 2 (PHP Runtime) PHP script

问题 The if statement is not running .It works fine with the tutorial I have followed but when i have implemented it in my code it fails.The error is Undefined index: thumbnailPic, The HTML code is : <label >Thumbnail Picture<text>*</text></label><br> <input type="file" name="thumbnailPic" id="pic"><br> <label >Original Picture<text>*</text></label><br> <input type="file" name="originalPic" id="pic"><br> The PHP code is : if (is_uploaded_file($_FILES['thumbnailPic']['tmp_name']) != false) { $spID=

问题 I've got an issue with turning a mysqli query's results into variables that I can populate text areas with. I've been trying to search for examples of this all around the web and I cannot seem to find something that resolves my issue. My last project along these lines used mysql, not mysqli, so it's not exactly the same, plus my project is varied enough to not allow for a code copy-paste. I'm trying to make a connection to a database, pull information used in a post element as a search

问题 This question already has an answer here : Call to undefined function mysqli_result::num_rows() (1 answer) Closed 5 years ago . Hi I am requesting help with my PHP. I checked the code through and I couldn't spot any errors in it. I'm not that great with PHP, and I hope any of you could help. It should return either false or true but, I get this error Fatal error: Call to undefined method mysqli::num_rows() Here is my code: $mysqli is my mysqli database connector. $email = $_SESSION["email"];

问题 Hi i need some help with this error I keep getting. It works fine everywhere else in the script except when I try to call it within this one function. The function I am calling is: getTotalServers Error: Fatal error: Call to a member function prepare() on a non-object in C:\xampp2\htdocs\runeloft\ss_sources\util.php on line 208 Calling in this function: function display_table($online, $where, $num_servers = null){ global $aaaa, $bbbb; $num_per_page = 30; $page = 1; // how many records per

问题 I have been struggling to extract the parent category posts from Wordpress. When i am trying to display the only parent category posts the wordpress query displaying the sub-categories posts as well. How can omit the sub-category posts. please help!!.. <? // Get the post ID $id = get_the_ID(); //get all the categories ( this function will return all categories for a post in an array) $category= get_the_category( $id ); if ($category->category_parent == 0) { //extract the first category from

问题 I'm having an issues with my code in php... mysqli_error() expects exactly 1 parameter, 0 given in on line 63. Im getting 8 of these errors. Here is line 63: --die('Error: ' . mysqli_error());-- Here is the function it is in: if($email_exists == false && $pass1 == $pass2) { $sql="INSERT INTO users(u_Fname, u_Lname, u_email, u_Password) VALUES('$Fname', '$Lname', '$EmailAddress', '$Password')"; if(!mysqli_query($link, $sql)) { #$query = mysqli_query($myConnection, $sqlCommand); #die (mysqli

问题 Suggested that we have this SQL text with 3 MySQL commands: show tables; some nonesense command which makes error; show tables; The first and third SQL are valid, the second isn't. So, when passing this SQL text to mysqli_multi_query, does MySQL stop execution at the second command (error one) or it tries to go on until the last command? Coz it is dangerous if those commands after the error one relies on the error one. 回答1: No, if one query fails the rest will not be executed. See this