Display Entry From Database To Browser
I can't display entry from database to browser, what's wrong with my code? Thank you for your kind attention. Code:
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If I want to enter some info into 2 database's with some fields being the same eg name and pass and the key 'uid' to be the same i got this code and it give me a error, PHP Code:
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I need to change an entry in a database and I don't know how...they said..i should use update..but still ..it's all too vague for me... See, I need to modify a function which deletes ads. The ads are grouped in one table aptly named "ads". Now, in a seperate table "category" is where I have saved the total number of ads in a record aptly titled = "total". These two tables have two things in common, they are joined by the category number=catid Now, my problem is that whenever somebody deletes an ad. It just does that. It deletes the ad in table "ads". Now I need to change the total number of ads ("total") in table "category". Please do tell me how do i do this?
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I am developing a regisration page that will present a set of check boxes to the viewer. This list of checkboxes is developed from a list in a database so the amount and names of the boxes will change. This portion works just fine. The page content has a user name and password along with several sets of these checkboxes. All the form content of this page is presented in one form. The problem: I want to store the values presented in these checkboxes back into the database based on what the user selects. The data will be stored in a table that has the id for the user and the id for the checkbox item. As anyone knows, checkbox's only send a boolean value through POST so once the data is sent it is nearly impossible to test for its value and work on it since these are dynamically presented. What I mean by dynamically presented is that the list is pulled from a database. The checkboxes are build as such: <p><input type="checkbox" name="{$row['name']}">{$row['name']}</p>. $row is the reference to the database result set that has been affected by the row_assoc function. Since I have no idea what the contents of $row['name'] will be. I cannot use isset($_POST['checkboxname']) to test its value as you would normally do. As I see it 'checkboxname' is just a string and so it seems logical to me that this would work: isset($_POST['$row['name']']), but it doesn't. My request: I need to be able to check to see if the checkbox is check like such: isset($_POST['$row['name']']) so I can check the value of $_POST['$row['name']']. What is the correct way to run this check if it is at all possible? Or is there another way to do this or a more effective way to do this?
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I have got an assignment which has to display the FAQ section somewhat like shown in the pasted URL's FAQ section. Please help me in figured out the code to display the content in the same format. Code:
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I created a very simple form and a MySQL database to see if I could get the information submitted from the form to the database, and also a copy of the form submission emailed to me. I have managed to do that with the help of a few online tutorials. Now, I would like to collect the IP address and browser type of the person submitting the form and have this information transferred to my MySQL database, along with the other form field information. I would like this information to be submitted to columns in my MySQL database labeled "ip" and "browser" respectively. Code:
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I just wanted to make something that creates html-code which should be inserted in a mysql database. The user should type text in a textarea. When the user submits the data, all enters should be translated into "<br>" and it would be nice if the user could also insert links ( "<a href="..." ) in the formular. Does someone knows a good, very simple script? There are wysiwyg editors on the net, but I don't need to modify fonts, colors et cetera. The most important thing are the "<br>" tags which should be generated.
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I have been struggling with this problem for a while and using search extensively has not been very fruitful, so I decided to post this here. I am making a database driven image gallery and practically everything works - I upload a jpeg as binary data to a database and can display it in a browser. However, I intend for people to be able to point to the jpeg image and use the "save as" option to save the image as a jpeg. The jpegs are high quality and quite large, since it is intended for the press to be able to use it in their publications. However, when I right-click on the image and choose "save as", the only option that IE gives me is the .bmp format. Opera, on the other hand, tries to save it as a .phtml file that I use to output the image. It does work if I manually force Opera to save it as a JPEG, but IE will only save it as a .bmp. Does anyone know how to tell the browser that it's a jpeg image?
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I am calling all fields from a table, just to many to call the ones i need. I would like to drop the first 2 from being shown, how can I do that by the code below or do I need to rewrite completely? while ($line = mysql_fetch_array($store_data, MYSQL_ASSOC)) { foreach ($line as $data) { echo $data . ', '; } }
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i have a php script that read from mysql and display then in textbox. I have problem with the way textbox display the data. <input type=text name="package_name" size=20 maxlength=10 value='.$package['package_name'].'> when the package_name is 'china tour' it will only show china in the text box. How can i show all the content of the record?
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I am new to PHP and Mysql but am really determined to learn these techniques. I have some script that reads and displays results from the databse onto a page. However, I really wish to limit the number of results that are displayed to 10 per page and then have the next/privious links under the results. With these results, the results under the email field should be hyperlinked so that the person viewing the results should double click on the email link and send the person an email. How can i do that? I have to learn this technique. The code is:-
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this functions adds new locations in am_ups_shipment_location table after a tag in XML is detected it is pointed towards this function and this function then goes and adds the new location. for the first time when i run the php prgoram the new location gets added.but when i refresh ,again the same location gets added but with different id[id is on autoincrement]. here is how my table looks after refresh.[you can see duplicate entry].This is what i dont wnat...i dont want a duplicate entry. Can you please help me get rid of this duplicate entry by writing some update functions and checking against already present data in the table? Code:
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I'm creating a news database where I'd like to display each article Headline along with the first, say 300 characters of the article. At it stands, it's displayiong the full article in the table. PHP Code:
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Random Entry From Mysql
I am trying to select a random entry from mysql. Anyway - My version of mysql doesn't have the built in rand() function, so I am using PHP's. Anyway - I wrote the following function... function displayguide(){ $rand = rand(10,202000); connect(); $query = "SELECT *, $rand as random FROM guide_rotater ORDER BY random LIMIT 1"; $result = mysql_query($query); $this = mysql_fetch_object($result); print "<b>$this->title</b><p>$this->description<p><a href="$this->url"><b>Read the guide</b></a>";} Anyway - it works, the only problem is that it uses the same $rand number for all of them so it isn't random I can't think of a way to do this using php's rand function.
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