skidmark10 Posted August 15, 2011 Share Posted 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! Quote Link to comment https://forums.phpfreaks.com/topic/244862-select-mysql-display-in-an-html-table-using-php/ Share on other sites More sharing options...
jcbones Posted August 15, 2011 Share Posted 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); ?> Quote Link to comment https://forums.phpfreaks.com/topic/244862-select-mysql-display-in-an-html-table-using-php/#findComment-1257853 Share on other sites More sharing options...
skidmark10 Posted August 15, 2011 Author Share Posted 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 ';' Quote Link to comment https://forums.phpfreaks.com/topic/244862-select-mysql-display-in-an-html-table-using-php/#findComment-1257864 Share on other sites More sharing options...
jcbones Posted August 15, 2011 Share Posted August 15, 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.) Quote Link to comment https://forums.phpfreaks.com/topic/244862-select-mysql-display-in-an-html-table-using-php/#findComment-1257874 Share on other sites More sharing options...
skidmark10 Posted August 16, 2011 Author Share Posted 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>"; Quote Link to comment https://forums.phpfreaks.com/topic/244862-select-mysql-display-in-an-html-table-using-php/#findComment-1257958 Share on other sites More sharing options...
phpSensei Posted August 16, 2011 Share Posted August 16, 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); ?> Quote Link to comment https://forums.phpfreaks.com/topic/244862-select-mysql-display-in-an-html-table-using-php/#findComment-1257965 Share on other sites More sharing options...
jcbones Posted August 16, 2011 Share Posted August 16, 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! Quote Link to comment https://forums.phpfreaks.com/topic/244862-select-mysql-display-in-an-html-table-using-php/#findComment-1258309 Share on other sites More sharing options...
phpSensei Posted August 16, 2011 Share Posted August 16, 2011 I was being mean to people? Quote Link to comment https://forums.phpfreaks.com/topic/244862-select-mysql-display-in-an-html-table-using-php/#findComment-1258317 Share on other sites More sharing options...
skidmark10 Posted August 17, 2011 Author Share Posted 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! Quote Link to comment https://forums.phpfreaks.com/topic/244862-select-mysql-display-in-an-html-table-using-php/#findComment-1258368 Share on other sites More sharing options...
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