IE9 can’t open the php page, please help me, what should I do?
IE9 can’t open the php page, please help.I write code in WAMP environment, and every time IE9 runs the php page, a download dialog pops up, but I can parse it normally with 360 browser! In addition, my IE9 always pops up a download dialog box for all other PHP pages on the Internet, but the 360 can be displayed normally! What is the reason for this? I can’t find the reason after searching online. Can anyone help me? ? ? ? ? ? ——Solution——————–Content is not configured -type——Solution——————–Firefox How about next? How to access? Have the php files been placed in the same directory? ——Solution——————–Last of this post Edited by xuzuning on 2012-10-06 11:08:35 For the url that always pops up a download dialog box when using IE9 You execute the following code: $s = file_get_contents(‘real url’); echo base64_encode(substr($s, 0, 100)); Post results Or give a valid link ——Solution——————–I know what’s going on, The php.ini file configuration is incorrect, and the part of the dll file that loads the php extension is incorrect. Sir, you should check the PHP operating environment.
PHPeclipse configuration related issues, please help me
PHP eclipse configuration problem, please help me. Today I spent an afternoon working on php configuration issues, so let’s get right to the point. I am eclipse + wamp + zDebug Everything was done according to the step-by-step instructions on the Internet. But finally right click on the .php file–>run as php application At this time, another window will appear, displaying http://localhoast…but as mentioned, counld not display webpage. My configuration diagram is like this: Then in windows –> properties That zend debug.dll file, I put it in php under the wamp file. The path is correct. Please help me take a look. Where does the problem occur? ? Thank you all. ——Solution——————–Use zend directly studio bar——Solution——————–dreamweaver The most comfortable one~~The eclipse 4 php is not very easy to use. Personal feeling
PHP pseudo-static, please help me.
php pseudo-static, please help meRewriteRule ^img/([0-9]+).html?p=(.*)$ img.php?id-$1.html&p=$2 Opinions that are not pseudo-static can be accessed The effect I want 123.com/img/6.html?p=1 I think I got it right. Why prompt 404 ——Solution——————–RewriteRule ^img/([0-9]+).html?p=([1-9]+)$ img.php?id-$1.html&p=$2
Please help me, I need answers tonight! ! ! _html/css_WEB-ITnose-html tutorial
Employee query system 1. Language and environment A. Implementation language: Java B. Implementation technology: HTML, Javascript, JSP, Servlet C. Environmental requirements : Eclipse, database (Oracle, MySql optional), Tomcat D. No style requirements when designing web pages 2. Requirements XXX company has a personnel management system, one of which is a functional module It is to query the details of employees based on their positions and names. 2.1 Function and page design requirements: 1) The query page can display the employee’s position name. The display content of the position name comes from the database table Employee. Job titles cannot be repeated. 2) The query page can complete the client verification work and provide verification prompts for unselected job categories. 3) The result display page can query the results based on the entered employee name and selected employee position, and correctly use the table to display the results. If the query content is empty, the prompt message should be displayed correctly. 2.2 Database design requirements: 1) All fields of the database table Employee must be set according to the contents of (Table 1). The content is filled in the data table according to the content of (Table 2). Employee Table 1: Table…
Please help me with PHP problem, please help me.
//refresh function//refresh item statusfunction refresh(){ $problem = false; $result = mysql_query(“SELECT ITEM_ID FROM ITEM WHERE END > NOW() AND STATUS ! = ‘SOLD'”); if($result) { while ($row = mysql_fetch_array($result)) { $act_id = $row[‘ITEM_ID’]; mysql_query(“UPDATE ITEM SET STATUS = ‘ACTIVE’ WHERE ITEM_ID = $act_id”); } $result = mysql_query(“SELECT ITEM_ID, WINNER_ID, END FROM ITEM WHERE END <= NOW() AND STATUS != 'SOLD'"); if($result) { while ($row = mysql_fetch_array( $result)) { $exp_id = $row['ITEM_ID']; $win_id = $row['WINNER_ID']; $time = $row['END']; // Error here. There is only one row has been displayed. There should be 4 rows actully. // Result: 7 0 2013-04-14 00:00:00 echo $exp_id."”; echo $win_id.””; echo $time.” “; if($win_id == 0) { mysql_query(“UPDATE ITEM SET STATUS = ‘EXPIRE’ WHERE ITEM_ID = $exp_id”); } else { mysql_query(“UPDATE ITEM SET STATUS = ‘SOLD’ WHERE ITEM_ID = $exp_id”); echo “INSERT INTO ITEM_SOLD VALUES($exp_id,$win_id,'”.$time.”‘,’N’)”; mysql_query(“INSERT INTO ITEM_SOLD VALUES($exp_id,$win_id,'”. $time.”‘,’N’)”); } } $result = mysql_query(“SELECT ITEM_ID FROM ITEM WHERE BEGIN > NOW() AND STATUS != ‘SOLD'”); if($result) { while ($ row = mysql_fetch_array($result)) { $na_id = $row[‘ITEM_ID’]; mysql_query(“UPDATE ITEM SET STATUS = ‘NA’ WHERE ITEM_ID = $na_id”); } } else { $problem = true; } } else { $problem = true; } } else { $problem…
I am new to PHP, please help me.
I am engaged in Linux server maintenance and C development, and I am relatively familiar with Linux. Now I want to develop into PHP and build a simple web access server. First of all, I have never been exposed to PHP. Now I would like to ask the master how to get started. Are there any tutorials or anything like that? Reply to discussion (solution) I am also a newbie in PHP and don’t know anything Website environment construction《《 Newbie tutorial《《》 Is it difficult? How long does it take to create a simple web page? The master is in the PHP area, not in the water area How simple There is still one less thing to do; How simple It’s a good choice to watch basic videos There’s still a lot to be done; How simple
