skidmark10

August 23, 2011

Hello,

Does anyone know the php code to display mysql table names?

August 18, 2011

You mean a JOIN?

Yes I mean "JOIN". Sorry I am new to this.

August 18, 2011

Hello,

I am trying to select more than two mysql tables using php. All of the tables have one column in common "Client_ID". Does anyone know how I would pull information from all the columns with the same "Client_ID"?

I don't have a code I'm working with yet, I'm stumped.

August 17, 2011

this line

. $row['LastName']"</td>";

Is missing a '.'

try

<?php
mysql_select_db("my_db", $con);
$result = mysql_query("SELECT * FROM Persons");
echo "<table border='1'><tr><th>Name</th></tr>";
while($row = mysql_fetch_array($result))  {  
echo "<tr>";  echo "<td>" . $row['FirstName'] . ' ' . $row['LastName']. "</td>";  
echo "</tr>";  
}
echo "</table>";
mysql_close($con);
?>

I apologize for that, I'm gonna have to get more sleep, and quit being mean to people!

Thank you so much!

August 16, 2011

No where in your first post did you say anything about a parse error! Post your ENTIRE script, that is the only way to deal with parse errors. (Include the acutal error also, there should be a line number in it.)

I tried a similar code (not the code I posted)

I put this and got a parse error

echo "<td>" . $row['FirstName'] . " . $row['LastName']"</td>"

;

Then I tried your code too and received a parse error

echo "<td>" . $row['FirstName'] . ' ' . $row['Last_Name']"</td>";

August 15, 2011

mysql_select_db("my_db", $con);
$result = mysql_query("SELECT * FROM Persons");
echo "<table border='1'><tr><th>Name</th></tr>";
while($row = mysql_fetch_array($result))  {  
echo "<tr>";  echo "<td>" . $row['FirstName'] . ' ' . $row['LastName']"</td>";  
echo "</tr>";  
}
echo "</table>";
mysql_close($con);
?>

I don't think that works, I keep receiving the same error as before :-\

Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';'

August 15, 2011

Hello,

I am trying to retrieve and display data from a mysql table into a html table using php. I am using this code and it works successfully.

<?php
mysql_select_db("my_db", $con);

$result = mysql_query("SELECT * FROM Persons");

echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['FirstName'] . "</td>";
  echo "<td>" . $row['LastName'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?>

However, I want $row['FirstName'] and $row['LastName'] to appear in the SAME cell of the html table, next to one another. How would I write the code for that, if it is possible

Thanks in advance!

August 8, 2011

What do you mean you're using phpMyAdmin? How does a user enter anything in your database through phpMyAdmin?

I'm sorry, forgot to mention that I have an html form that posts data into the mysql table. phpmyadmin is what I'm using for my database.

August 8, 2011

Hello,

I am using an html form where users submit their date of birth into a column in my mysql table. I am using phpmyadmin. I also have a table for current age that should get the users current age based on the information they submitted into the "date of birth" column.

What date function would allow me to do the above? Also, where would I put that function? Again, I am using phpmyadmin.

August 1, 2011

Hello,

I have an html form with checkbox answers that I want to store into a mysql database. When I use php however, only one of the values is stored in the column. I know this isn't a mysql issue because my table is "SET".

Any reason why it isn't working?

August 1, 2011

I have an html form with checkbox answers. I want to store the answers in ONE mysql column. When I select for checkbox answers, only the very last one shows up in the column, but I have it enumerated so I don't know why it shows up like that.

Any suggestions?

July 31, 2011

Fixed it

Instead of

$sql = "SELECT * FROM `Partners_Body_Types` WHERE Client_ID=" .$_POST['clientid']. ""or die ('Error: '.mysql_error ());

It should be

$sql = "SELECT * FROM `Partners_Body_Types` WHERE Client_ID='$_POST[clientid]'"

July 31, 2011

The query is failing for some reason, but you have no error-handling logic in place to catch that and report any errors. Is that all of the code, or is there more to it?

How can I put in the error-handling logic?

This is all of my php code (minus my database connection code), I'll post my html code so you can see the form.

HTML

<form name="input" action="fetchclientprofile.php" method="GET">
<center> Client ID <input type="text" name="clientid" /><br /> </center>
<center> <input type="submit" value="Retrieve" /> </center>

PHP

$sql = "SELECT * FROM `Partners_Body_Types` WHERE Client_ID= " .$_POST['clientid']. "";

$query = mysql_query($sql);

echo "<table border='0'>

<tr>

<th>Client ID</th>

<th>Partner's Body Type</th>

</tr>";

while ($row = mysql_fetch_array($query))

{

echo "<tr>";

echo "<td>" . $row['Client_ID'] . "</td>";

echo "<td>" . $row['Slim_Slender'] . "</td>";

echo "<td>" . $row['Average'] . "</td>";

echo "<td>" . $row['Athletic'] . "</td>";

echo "<td>" . $row['Heavy'] . "</td>";

echo "</tr>";

}

echo "</table>";

mysql_close($con);

?>

July 31, 2011

please mark as solved. the topic solved button can be found at the bottom left of the page

I am still receiving the error Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource after making the corrections. Is there anything else I can do?

$sql = "SELECT * FROM `Partners_Body_Types` WHERE Client_ID= " .$_POST['clientid']. "";

$query = mysql_query($sql);

echo "<table border='0'>
<tr>
<th>Client ID</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
</tr>";

while ($row = mysql_fetch_array($query))
   {
   echo "<tr>";
   echo "<td>" . $row['Client_ID'] . "</td>";
   echo "<td>" . $row['Slim_Slender'] . "</td>";
   echo "<td>" . $row['Average'] . "</td>";
   echo "<td>" . $row['Athletic'] . "</td>";
   echo "<td>" . $row['Heavy'] . "</td>";
   echo "</tr>";
   }
echo "</table>";

mysql_close($con);
?>

July 31, 2011

I've never encountered this error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

Does anyone knows where it stems from, and ways to fix it?

Here's my code just in case.

$sql = "SELECT * FROM `Partners_Body_Types` WHERE Client_ID=" .$_POST['clientid']. ""or die ('Error: '.mysql_error ());

$query = mysql_query($sql);

echo "<table border='0'>
<tr>
<th>Client ID</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
</tr>";

while ($row = mysql_fetch_array($sql));
   {
   echo "<tr>";
   echo "<td>" . $row['Client_ID'] . "</td>";
   echo "<td>" . $row['Slim_Slender'] . "</td>";
   echo "<td>" . $row['Average'] . "</td>";
   echo "<td>" . $row['Athletic'] . "</td>";
   echo "<td>" . $row['Heavy'] . "</td>";
   echo "</tr>";
   }
echo "</table>";

mysql_close($con);
?>

July 31, 2011

Thank you

July 31, 2011

I keep recieving the error: Parse error: syntax error, unexpected T_ECHO on Line 13. Does anyone know what "T_ECHO" means, so I can know for future reference?

Here's my code, Line 13 is echo "<table border='0'>.

Thanks in advance!

$sql = "SELECT * FROM Partners_Body_Types WHERE Client_ID=" .$_POST['clientid']. "";
$query = mysql_query($sql)

echo "<table border='0'>
<tr>
<th>Client ID</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
</tr>";

July 31, 2011

Post the code you've written, and describe the problems you're having with it.

Really haven't started any code yet, I'm confused.

I know how to create an html form, with the form action of "post", and it stores the form data into the mysql table.

Now I want an html form where I enter a client's ID number, hit submit, and the row with the ID number in the table is displayed.

July 30, 2011

Hi,

I have a table in mysql that I would like to retrieve data from, but only a particular row. I want to have an html form where I put the client id number in, hit submit, and have that member's row of data display (one row in the mysql table).

Does anyone know the code for this?

July 30, 2011

Thank YOU

July 30, 2011

I am trying to insert answers from a checkbox on an html form into a mysql table using php. For some odd reason, the data doesnt make it to the mysql table. I'm not sure what the issue is, so I posted all of the information I have so I can see where I made my mistake.

thanks in advance.

Here is the html

<table class=MsoNormalTable border=0 cellpadding=0 width="100%"
style='width:100.0%;mso-cellspacing:1.5pt;mso-yfti-tbllook:1184;mso-padding-alt:
0in 5.4pt 0in 5.4pt'>
<tr style='mso-yfti-irow:0;mso-yfti-firstrow:yes'>
  <td width="100%" style='width:100.0%;padding:.75pt .75pt .75pt .75pt'>
  <p class=MsoNormal><span style='mso-fareast-font-family:"Times New Roman"'><INPUT 

TYPE="checkbox" NAME="partnerbodytype1" VALUE="Slender|Slim"> Slender/Slim
  <o:p></o:p></span></p>
  </td>
</tr>
<tr style='mso-yfti-irow:1'>
  <td width="100%" style='width:100.0%;padding:.75pt .75pt .75pt .75pt'>
  <p class=MsoNormal><span style='mso-fareast-font-family:"Times New Roman"'><INPUT 

TYPE="checkbox" NAME="partnerbodytype2" VALUE="Average"> Average
  <o:p></o:p></span></p>
  </td>
</tr>
<tr style='mso-yfti-irow:2'>
  <td width="100%" style='width:100.0%;padding:.75pt .75pt .75pt .75pt'>
  <p class=MsoNormal><span style='mso-fareast-font-family:"Times New Roman"'><INPUT 

TYPE="checkbox" NAME="partnerbodytype3" VALUE="Athletic"> Athletic
  <o:p></o:p></span></p>
  </td>
</tr>
<tr style='mso-yfti-irow:3;mso-yfti-lastrow:yes'>
  <td width="100%" style='width:100.0%;padding:.75pt .75pt .75pt .75pt'>
  <p class=MsoNormal><span style='mso-fareast-font-family:"Times New Roman"'><INPUT 

TYPE="checkbox" NAME="partnerbodytype4" VALUE="Heavy"> Heavy
  <o:p></o:p></span></p>
  </td>
</tr>
</table>

</div>

Here is the php code

$sqlpartnersbodytype= "INSERT INTO `Partners_Body_Types` (Slim_Slender, Average, Athletic, Heavy) VALUES ('$_POST[partnerbodytype1]','$_POST[partnerbodytype2]',
'$_POST[partnerbodytype3]','$_POST[partnerbodytype4]')";

Here is the table information

 `Partners_Body_Types` CHANGE  `Slim_Slender`  `Slim_Slender` VARCHAR( 10 ) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL ,
CHANGE  `Average`  `Average` VARCHAR( 10 ) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL ,
CHANGE  `Athletic`  `Athletic` VARCHAR( 10 ) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL ,
CHANGE  `Heavy`  `Heavy` VARCHAR( 10 ) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL

July 30, 2011

what do you currently have it set as in the database?

CHANGE `Slim_Slender` `Slim_Slender` ENUM( 'Slim|Slender' ) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL ,

CHANGE `Average` `Average` ENUM( 'Average' ) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL ,

CHANGE `Athletic` `Athletic` ENUM( 'Athletic' ) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL ,

CHANGE `Heavy` `Heavy` ENUM( 'Heavy' ) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL

July 30, 2011

Hello,

I am trying to store checkbox answers into a table in mysql. What should the table input be? I'm using enumeration, but my answers aren't appearing. What am I doing wrong?

July 30, 2011

You could assign an individual name to each value... make sure to update your database table as well:

<td><div align="left">Please select all body types you would be interested in dating: </div></td>

<td><input name="partnerbodytype1" type="checkbox" value="Slender" /> Slenders

<input name="partnerbodytype2" type="checkbox" value="Average" /> Average<br />

<input name="partnerbodytype3" type="checkbox" value="Athletic" /> Athletic

<input name="partnerbodytype4" type="checkbox" value="Heavy" /> Heavy<br />

php:

$partnersbodytype1 = $_POST['partnersbodytype1']

$partnersbodytype2 = $_POST['partnersbodytype2']

$partnersbodytype3 = $_POST['partnersbodytype3']

$partnersbodytype4 = $_POST['partnersbodytype4']

$sqlClientInfo= "INSERT INTO `Client_Info` (partnersbodytype1, partnersbodytype2, partnersbodytype3, partnersbodytype4)

VALUES ('$partnersbodytype1','$partnersbodytype2','$partnersbodytype3','$partnersbodytype4')

I've tried this way, but my answers still don't appear. Is there something wrong with my table? Is there something I can change with the table?

July 29, 2011

You could assign an individual name to each value... make sure to update your database table as well:

<td><div align="left">Please select all body types you would be interested in dating: </div></td>

<td><input name="partnerbodytype1" type="checkbox" value="Slender" /> Slenders

<input name="partnerbodytype2" type="checkbox" value="Average" /> Average<br />

<input name="partnerbodytype3" type="checkbox" value="Athletic" /> Athletic

<input name="partnerbodytype4" type="checkbox" value="Heavy" /> Heavy<br />

php:

$partnersbodytype1 = $_POST['partnersbodytype1']

$partnersbodytype2 = $_POST['partnersbodytype2']

$partnersbodytype3 = $_POST['partnersbodytype3']

$partnersbodytype4 = $_POST['partnersbodytype4']

$sqlClientInfo= "INSERT INTO `Client_Info` (partnersbodytype1, partnersbodytype2, partnersbodytype3, partnersbodytype4)

VALUES ('$partnersbodytype1','$partnersbodytype2','$partnersbodytype3','$partnersbodytype4')

Yikes! Is this the only way???? If so, I have to prepare myself for a LONG night of work....

Sign In

skidmark10

Posts

Joined

Last visited

Content Type

Profiles

Forums

Posts posted by skidmark10

How to display column names in a mysql table using php

Selecting more than two mysql tables tables using php?

Selecting more than two mysql tables tables using php?

Select mysql & display in an html table using php

Select mysql & display in an html table using php

Select mysql & display in an html table using php

Select mysql & display in an html table using php

Current Age and Date function? phpmyadmin

Current Age and Date function? phpmyadmin

Pass multiple checkbox answers from html into a mysql table using php

How Do I Store Multiple Answers in ONE column?

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

Does anyone knows what: "Parse error: syntax error, unexpected T_ECHO" means?

Does anyone knows what: "Parse error: syntax error, unexpected T_ECHO" means?

How to display data from mysql using an HTML form?

How to display data from mysql using an HTML form?

Error: storing things from an html form into a mysql table

Error: storing things from an html form into a mysql table

Storing Checkbox answers from an HTML form into mysql

Storing Checkbox answers from an HTML form into mysql

How to store checkbox answers from html form into mysql?

How to store checkbox answers from html form into mysql?

Browse

Activity

Important Information