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Hey, I'm getting a syntax error with a php file I'm creating to make my database searchable:

[code]
$query = "SELECT * FROM 'table01' NATURAL JOIN 'ID' WHERE 'table01'.'Name' like '%$qstring%'";
[/code]

Is this outdated syntax?  I'm using PHP 5.  Tried googling with little success.  Thanks for any help.
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Hmmm...I'm not sure if this is doing what I need it to do...I want the user to be able to search for strings for fields like 'Name,' and this code will query the database for the ID (primary key) of any records with that 'Name', and display the records.  Is there a query that would work better?  Thanks!  :)
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https://forums.phpfreaks.com/topic/23027-syntax-error/#findComment-104402
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Try this.
[code]
<?php

$request = mysql_query("SELECT ID, Name FROM tableName WHERE colName LIKE '%$searchTerm%'");

echo '
<table>';

// Loop through the results.
while ($row = mysql_fetch_assoc($request))
echo '
<tr>
<td>', $row['ID'], '</td>
<td>', $row['Name'], '</td>
</tr>';

// free the results
mysql_free_result($request);

echo '
</table>';
?>
[/code]
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https://forums.phpfreaks.com/topic/23027-syntax-error/#findComment-104410
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[code]
<?php
include('db_login.php');
$connection = mysql_connect($db_host, $db_username, $db_password);
$db_select = mysql_select_db($my_database);

    $db = "my_database";
    mysql_select_db($db) or die("Could not select the database '" . $db . "'.  Are you sure it exists?");
if (!$connection)
{
die ("Could not connect to the database: <br />". mysql_error());
}

$request = mysql_query("SELECT ID, Name FROM tableName WHERE colName LIKE '%$searchTerm%'");

echo '
<table>';

// Loop through the results.
while ($row = mysql_fetch_assoc($request))
echo '
<tr>
<td>', $row['ID'], '</td>
<td>', $row['Name'], '</td>
</tr>';

// free the results
mysql_free_result($request);

echo '
</table>';
}
// Close the connection
mysql_close($connection);
?>
[/code]
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https://forums.phpfreaks.com/topic/23027-syntax-error/#findComment-104459
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Om here you go.  There was a } after the closing table tag.
[code]
<?php
include('db_login.php');
$connection = mysql_connect($db_host, $db_username, $db_password);
$db_select = mysql_select_db($my_database);

    $db = "my_database";
    mysql_select_db($db) or die("Could not select the database '" . $db . "'.  Are you sure it exists?");
if (!$connection)
die ("Could not connect to the database: <br />". mysql_error());

$request = mysql_query("SELECT ID, Name FROM tableName WHERE colName LIKE '%$searchTerm%'");

echo '
<table>';

// Loop through the results.
while ($row = mysql_fetch_assoc($request))
echo '
<tr>
<td>', $row['ID'], '</td>
<td>', $row['Name'], '</td>
</tr>';

// free the results
mysql_free_result($request);

echo '
</table>';

// Close the connection
mysql_close($connection);
?>
[/code]
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https://forums.phpfreaks.com/topic/23027-syntax-error/#findComment-104463
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Right now I have this:

[code]
$request = mysql_query("SELECT ID, Name, Organization, Title, Street, City, State, Zip FROM table01 WHERE ID LIKE '%$Name%'");
[/code]

I want the search string to be someone's Name, and the code to find the IDs (primary key) in my database that correspond with that name and display the available records...
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https://forums.phpfreaks.com/topic/23027-syntax-error/#findComment-104483
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Yes, I'm using a form which takes the 'Name' string from the user and passes it to a php file that's supposed to find the IDs in the database that correspond to that name and display the records.  Thanks, I tried $_POST['Name'] but still am getting a blank result when I try to search for a Name.
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https://forums.phpfreaks.com/topic/23027-syntax-error/#findComment-104503
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Sure...here's my form:

[code]
<p><b><font color="blue">Search Records:</b><br /></font>
<form action="search.php" method="post" >
Name:<br /> <input type="text" name="Name" /><br />
<input type="submit" value="Submit" />
<input type="reset" value="Reset">
</form></p>
[/code]


and here's the php file:

[code]
<?php
include('db_login.php');
$connection = mysql_connect($db_host, $db_username, $db_password);
$db_select = mysql_select_db($my_database);

    $db = "my_database";
    mysql_select_db($db) or die("Could not select the database '" . $db . "'.  Are you sure it exists?");
if (!$connection)
{
die ("Could not connect to the database: <br />". mysql_error());
}

$request = mysql_query("SELECT ID, Name, Organization, Title, Street, City, State, Zip FROM table01 WHERE ID LIKE '%$_POST['Name']%'");

// Loop through the results.
while ($result_row = mysql_fetch_row(($result)))
{
echo 'ID: '.$result_row[0] . '<br />';
echo 'Name: '.$result_row[1] . '<br />';
echo 'Organization: '.$result_row[2]. '<br />';
echo 'Title: '.$result_row[3]. '<br />';
echo 'Street: '.$result_row[4] . '<br />';
echo 'City: '.$result_row[5]. '<br />';
echo 'State: '.$result_row[6]. '<br />';
echo 'Zip: '.$result_row[7]. '<br /><br />';
}
// free the results
mysql_free_result($request);


// Close the connection
mysql_close($connection);
?>
[/code]
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https://forums.phpfreaks.com/topic/23027-syntax-error/#findComment-104512
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You changed this line [code=php:0]$request = mysql_query("SELECT ID, Name, Organization, Title, Street, City, State, Zip FROM table01 WHERE ID LIKE '%$_POST['Name']%'");[/code] to this [code=php:0]$request = mysql_query("SELECT ID, Name, Organization, Title, Street, City, State, Zip FROM table01 WHERE ID LIKE '%$_POST[Name]%'");[/code]?
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https://forums.phpfreaks.com/topic/23027-syntax-error/#findComment-104546
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