the page code is all html and it wont load bec. of this error: Fatal error: Can't use function return value in write context line 142 code: <div> <div class=""> <input type="text" id="select_shelves" name="select_shelves" class="shelves_select_and_buttons" /> <div id="shelves_menu" > <ul> <li id="li_" oncli
该页面的代码是所有的HTML，它不会加载bec。 的这个错误： 致命错误：无法在写入上下文行142中使用函数返回值 码： <div> <div class=""> <input type="text" id="select_shelves" name="select_shelves" class="shelves_select_and_buttons" /> <div id="shelves_menu" > <ul> <li id="li_" onclick="printValue();">option5 <=
I'm getting this error and I can't make head or tail of it. The exact error message is: Fatal error: Can't use function return value in write context in /home/curricle/public_html/descarga/index.php on line 48 Line 48 is: if (isset($_POST('sms_code') == TRUE ) { Anybody knows what's going on??? PS Here's the full function, in case it helps: function validate_sms_cod
我得到这个错误，我不能做它的头或尾。 确切的错误信息是： 致命错误：无法在第48行的/home/curricle/public_html/descarga/index.php中的写入上下文中使用函数返回值 48行是： if (isset($_POST('sms_code') == TRUE ) { 任何人都知道发生了什么？ PS这是完整的功能，以防万一： function validate_sms_code() { $state = NOTHING_SUBMITED; if (isset($_POST('sms_code') == TRUE ) { $sms_code =
This question already has an answer here: PHP parse/syntax errors; and how to solve them? 13 answers function Passwordstuff ($pw, $pw2, $scpw, $pc, $pe) { $same = strcmp($pw, $pw2); if ($same == 0) { $scpw = hash("sha256", $pw, "true"); $pc = TRUE; $pe = 0; } else { $pc = FALSE; $pe = 1; } Missing closing } In case you are interested, have a look at a popular php style
这个问题在这里已经有了答案： PHP解析/语法错误; 以及如何解决它们？ 13个答案 function Passwordstuff ($pw, $pw2, $scpw, $pc, $pe) { $same = strcmp($pw, $pw2); if ($same == 0) { $scpw = hash("sha256", $pw, "true"); $pc = TRUE; $pe = 0; } else { $pc = FALSE; $pe = 1; } 缺少结束语} 如果你有兴趣，看看一个流行的php风格指南，以帮助提供一个“标准”的方式来格式化您的代码。 从长远
This question already has an answer here: PHP parse/syntax errors; and how to solve them? 13 answers Parse error: Syntax error, unexpected end of file in my PHP code 10 answers Could you replace <? endif ?> with <?php endif ?>
这个问题在这里已经有了答案： PHP解析/语法错误; 以及如何解决它们？ 13个答案 解析错误：语法错误，我的PHP代码中10个文件的意外结束答案 Could you replace <? endif ?> with <?php endif ?>
I got this error when debugging my code: PHP Parse error: syntax error, unexpected T_OBJECT_OPERATOR in order.php on line 72 Here is a snippet of the code (starting on line 72): $purchaseOrder = new PurchaseOrderFactory->instance(); $arrOrderDetails = $purchaseOrder->load($customerName); Unfortunately, it is not possible to call a method on an object just created with new before PHP 5
调试我的代码时出现此错误： PHP解析错误：语法错误，第72行的order.php中出现意外的T_OBJECT_OPERATOR 以下是代码片段（从第72行开始）： $purchaseOrder = new PurchaseOrderFactory->instance(); $arrOrderDetails = $purchaseOrder->load($customerName); 不幸的是，不可能在PHP 5.4之前的new对象上调用一个方法。 在PHP 5.4和更高版本中，可以使用以下内容： $purchaseOrder = (new PurchaseOrderFactory)-&g
This question already has an answer here: PHP parse/syntax errors; and how to solve them? 13 answers The last part of your code should read as follows: $mysqli="SELECT * FROM $tbl_name2 WHERE Emp_ID='$AdminI' and Emp_Firstname='$AdminU' and Emp_Lastname ='$AdminP'"; $result1=mysql_query($mysqli); $count1= mysql_num_rows($result1); $_SESSION1["login_val1"] = $count1; if($count==1) {
这个问题在这里已经有了答案： PHP解析/语法错误; 以及如何解决它们？ 13个答案 代码的最后一部分应该如下所示： $mysqli="SELECT * FROM $tbl_name2 WHERE Emp_ID='$AdminI' and Emp_Firstname='$AdminU' and Emp_Lastname ='$AdminP'"; $result1=mysql_query($mysqli); $count1= mysql_num_rows($result1); $_SESSION1["login_val1"] = $count1; if($count==1) { echo "login successful " . $_SESSION["login_
I'm getting an error: Parse error: syntax error, unexpected T_VARIABLE in home2/chippery/public_html/login/text/loginStuff.php on line 4 Line 4: $pass = $_POST['password'] However, line 3 works fine: $user = $_POST['username'] Post is from: <html> <form action='loginStuff.php' method='POST'> User: <input type='text' name='username'><br> Pas
我收到一个错误： 解析错误：语法错误，第4行中home2 / chippery / public_html / login / text / loginStuff.php中意外的T_VARIABLE 第4行： $ pass = $ _POST ['password'] 但是，第3行工作正常： $ user = $ _POST ['username'] 帖子来自： <html> <form action='loginStuff.php' method='POST'> User: <input type='text' name='username'><br> Pass: <input type='
Ok I have been asked by my college lecturer to create a login system using PHP. Now I read up about PHP seeing as Im not familiar to it. I used this code and coded some myself. <?php session_start(); $username = $_POST ['username']; $password = $_POST ['password']; if ($username&&$password) { $connect = mysql_connect("localhost","root","") or die("Couldn't Connect"); mysq
好的，我的大学讲师已经问过我用PHP创建一个登录系统。 现在我读了关于PHP看到我不熟悉它。 我使用这个代码并自己编码。 <?php session_start(); $username = $_POST ['username']; $password = $_POST ['password']; if ($username&&$password) { $connect = mysql_connect("localhost","root","") or die("Couldn't Connect"); mysql_select_db("phplogin") or die ("Couldn't find Database"); $
This question already has an answer here: Reference - What does this error mean in PHP? 30 answers You need to create a connection to your database first, do this using mysql_connect() like so: $con = mysql_connect(DATABASE, USERNAME, PASSWORD, TABLE); and then pass this connection whenever you make a query. Also change all of your function calls to mysqli_*
这个问题在这里已经有了答案： 参考 - 这个错误在PHP中意味着什么？ 30个答案 你需要首先创建一个到你的数据库的连接，使用mysql_connect()这样做： $con = mysql_connect(DATABASE, USERNAME, PASSWORD, TABLE); 然后在您进行查询时传递此连接。 还要将所有函数调用改为mysqli_ *
This question already has an answer here: mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource 31 answers try this code: foreach($data as $fl){ $result = @mysql_query(sprintf("SELECT * FROM useri WHERE fb_id='%s'", $fl["uid"])); if(is_resource($result) && mysql_num_rows($result) >= 1) { while ($row = my
这个问题在这里已经有了答案： mysql_fetch_array（）/ mysql_fetch_assoc（）/ mysql_fetch_row（）/ mysql_num_rows等...期望参数1是资源31答案 试试这个代码： foreach($data as $fl){ $result = @mysql_query(sprintf("SELECT * FROM useri WHERE fb_id='%s'", $fl["uid"])); if(is_resource($result) && mysql_num_rows($result) >= 1) { while ($row = mysql_fetch_array($result)) {