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Syntax Error On Line 32? Repair Immediately


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    Over the past few weeks, some readers have encountered a known syntax error on line 32. This issue occurs for a number of reasons. Let’s discuss some of them below.

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    syntax error at line 32

    A partner of mine once asked me when I can fix the power supply of his laptop, because I have

    Safe programmer

    Let my family take my laptop, oh-oh-oh-oh-oh-oh-oh-oh-oh, the problem is irreparable

    I am trying to create a login page with sessions that distinguish between 2 types of users Members: “Administrator” and “Subordinate”, with each type of employee being directed to a different incoming page after a successful login. This is actually accomplished by comparing usernames and comparing the password with the username and password in that database, with both fields matching, resulting in a successful login.

    Usernames, as well as the password entered by the user, are compared against values ​​from various tables that correspond to different user styles.

    However, I usually get a parsing error: syntax error, unexpected ”, ready for ‘(‘ on line 32. I did a little research and Stackoverflow, the next strategy I was able to find is related to an unexpected parsing error. ( It is expected in PHP when a content file is assigned a static property and an asset that doesn’t seem to belong to me based on what I can link to. Most of the load overflow issues I’ve seen regarding parsing errors are were you paying people whoSome people forget to close their customs “or”) “. However, I get a message because of an unexpected error about” {“while waiting for an internal (” “. Why is this so important ???

       0) // Username Found                    $ query = "SELECT * FROM users WHERE password = '$ _ post [password]'";           $ result2 = mysqli_query ($ cxn, $ query) or die ("Password not found");            $ num2 = mysqli_num_rows ($ result2);            if ($ num2> 0) // Password found safely                            $ _SESSION ['Auth'] = "Yes";                $ _SESSION ['Type'] = "Administrator";                title ("Location: AdminMain.php");                        another                            echo "Invalid username and / or password";                            elseif // Will be executed if the username is not found in the user table         <--- This should be a line informing me of an error  ke           $ query = "SELECT * FROM subuser WHERE username = '$ _ POST [username]'";            $ result3 = mysqli_query ($ cxn, $ query) or die ("Username not found");            $ num3 = mysqli_num_rows ($ result3);            if ($ num3> 0) // Username Found                           $ query = "SELECT * FROM USERS WHERE Password = '$ _ POST [Password]'";                $ result4 = mysqli_query ($ cxn, $ query);                $ num4 = mysqli_num_rows ($ result4);                if ($ num4> 0) // password found                                    $ _SESSION ['Auth'] = "Yes";                    $ _SESSION ['Type'] = "Less";                    title ("Location: SubMain.php");                                another                                    echo "Unable to find password";                                        another                            echo "Unable to get username";                            another                    echo "Invalid username and / or password entered";            include ("login page.php");    Pause; 

    I would probably appreciate it if anyone can take advantage of my error.

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