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Thanks .. that work.. -
Please rewrite this html line of code into php echo statement
Please write this html code into php echo statement.
<a href="javascrip t:ajaxpage('tes t.htm', 'contentarea'); ">test</a> -
link in CSS problem
I have a web page with css sections.
1. header
2. content
3. footer
Now i placed a menu in the header section and would like when a user click that link on the menu if will display in the content section. whenever the user clicks that page be displayed in the content section.
please advise... -
Drop down menu
My main web page has two frames. menu and content frame. i would like to a drop down ment to drop over the menu frame into the content frame when a mouse go over the menu.
How can i when my code look something like this.
Code:div id="menu_parent">Main Menu</div> <div id="menu_child"> <a href="#">Item 1</a> <a href="#">Item
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detect if a drop down list is seleted.
How can i detect if a user selected and use a Drop Down list in php so that i can update the change if it was selected, otherwise i do not update that field in table.
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Login php not working over network
I have login.php page where user entering the site must sign on. On the local machine it is working but over the network it not. i using fedora 8.
Any ideas? -
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Code:var xmlHttpfunction showUser(str) { xmlHttp=GetXmlHttpObject() if (xmlHttp==null) { alert ("Browser does not support HTTP Request") return } var url="getuser.php" url=url+"?q="+str url=url+"&sid="+Math.random() xmlHttp.onreadystatechange=stateChanged xmlHttp.open("GET",url,true) xmlHttp.send(null)
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DDL keep repeating.. when selected
My code is similar to the one below.
Code:<html> <head> <script src="selectuser.js"></script> </head> <body><form> Select a User: <select name="users" onchange="showUser(this.value)"> <option value="1">Peter Griffin</option> <option value="2">Lois Griffin</option>
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DDL Keep repeating.. on same page once selected.
My code is similar to the one below.
Code:<html> <head> <script src="selectuser.js"></script> </head> <body><form> Select a User: <select name="users" onchange="showUser(this.value)"> <option value="1">Peter Griffin</option> <option value="2">Lois Griffin</option>
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MYSQL default location problem.
I change the location of mysql. i placed it in a folder for example /uto/.
when i run my php application it not connecting to the database.. -
i getting results without the following statement.
Code:where e.equipmentLocationID = l.equipmentLocationID AND e.cabinetID = c.cabinetID AND e.equipmentOsID= osy.equipmentOsID AND e.equipmentServiceLevelID = s.equipmentServiceLevelID AND e.equipmentTypeID = eq.equipmentTypeID
whenever i include it returns nothing...Leave a comment:
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Code:connect(); // Build SQL Query $query = "select e.equipmentName, e.equipmentSerTagNum,l.location,c.name , osy.osName , s.type, e.hardwareDescription,e.macAddress,e.rackNum, eq.type,e.hostServerName,e.updatedDate,e.warrantyStartDate, e.warrantyEndDate ,e.equipmentBrand,e.equipmentApplication from equipmentInfo as e, equipmentLocation as l, cabinet as c, os as osy ,serviceLevelType as s,
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SQL statement help
i trying to read data from quite a few tables.. Now the statement works well except when i include the where e.equipmentLoca tionID = l.equipmentLoca tionID AND e.cabinetID = c.cabinetID AND e.equipmentOsID = osy.equipmentOs ID AND e.equipmentServ iceLevelID = s.equipmentServ iceLevelID AND e.equipmentType ID = eq.equipmentTyp eID . But i need to include since the output i want is not the ID but the field.
I am using PHP...
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princelindie replied to how do i get the id to store in my table instead of the name field displayed in DDLin PHP[php]print_r($_POST) ;[/php] this help.
was able to see where the error was..
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princelindie started a topic how do i get the id to store in my table instead of the name field displayed in DDLin PHPhow do i get the id to store in my table instead of the name field displayed in DDL
I am using code that looks something like this:
[php]
<?php
mysql_connect(" localhost", "xxxx_xx", "xxx") or die(mysql_error ());
mysql_select_db ("xxxx_xxx") or die(mysql_error ());
$query="SELECT Bloom_Name FROM ihs";
$result = mysql_query($qu ery);
echo '<select name="ihs" onchange="this. form.submit()"> '; ... -
Got it out.
Work fine.. what i did was point the link to _blank.Leave a comment:
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frameset problem
my site has a login page. Once you login you enter the main page which consist of frames: A top frame (menu frame) and a bottom frame(content frame).
now when the use click a menu option it displays in the content section that ok.
however when the user select logout it doing so with that frame instead of return to a full wide window .
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