Возможный дубликат:
PHP: Предупреждение: sort () ожидает, что параметр 1 будет массивом, ресурс указан
вот мой код: я не вижу ничего, что должно было вызывать это … идеи?
Предупреждение: mysql_fetch_array () ожидает, что параметр 1 будет ресурсом, boolean.
<?php include_once "mysql_connect.php"; if ($_POST['parse_var'] == "contactform"){ if(is_array($categories)) $whereCond = "in '".implode(",",$categories); else $wherecond = "= ".$categories; $dropdownValue = $_POST['dropdown']; $dropdownValue = mysql_real_escape_string($dropdownValue); $dropdownValue = eregi_replace("`", "", $dropdownValue); $searchField= $_POST['searchinput']; $searchField = mysql_real_escape_string($searchField); $searchField = eregi_replace("`", "", $searchField); if ($dropdownValue == "phone"){ $sql = mysql_query("SELECT * FROM pcparts WHERE phone='$searchField'"); while($row1 = mysql_fetch_array($sql)){ $arrayuserinfo[] = array( 'phone' => $row1["phone"], 'name' => $row1["name"], 'city' => $row1["city"], 'state' => $row1["state"], 'address' => $row1["address"], 'zip' => $row1["zip"], ); }; for($i=0;$i < count($arrayuserinfo);$i++){ $phone = $arrayuserinfo[$i]["phone"]; $name = $arrayuserinfo[$i]["name"]; $city = $arrayuserinfo[$i]["city"]; $state = $arrayuserinfo[$i]["state"]; $address = $arrayuserinfo[$i]["address"]; $zip = $arrayuserinfo[$i]["zip"]; echo"<table width='400' border='1' cellpadding='3'> <tr> <td>phone</td> <td>name</td> <td>address</td> <td>city</td> <td>state</td> <td>zip</td> </tr> <tr> <td>$phone</td> <td>$name</td> <td>$address</td> <td>$city</td> <td>$state</td> <td>$zip</td> </tr> </table><br /> "; } } else if ($dropdownValue == "name"){ $sql = mysql_query("SELECT * FROM pcparts WHERE name='$searchField'"); while($row1 = mysql_fetch_array($sql)){ $arrayuserinfo[] = array( 'phone' => $row1["phone"], 'name' => $row1["name"], 'city' => $row1["city"], 'state' => $row1["state"], 'address' => $row1["address"], 'zip' => $row1["zip"], ); }; for($i=0;$i < count($arrayuserinfo);$i++){ $phone = $arrayuserinfo[$i]["phone"]; $name = $arrayuserinfo[$i]["name"]; $city = $arrayuserinfo[$i]["city"]; $state = $arrayuserinfo[$i]["state"]; $address = $arrayuserinfo[$i]["address"]; $zip = $arrayuserinfo[$i]["zip"]; echo"<table width='400' border='1' cellpadding='3'> <tr> <td>phone</td> <td>name</td> <td>address</td> <td>city</td> <td>state</td> <td>zip</td> </tr> <tr> <td>$phone</td> <td>$name</td> <td>$address</td> <td>$city</td> <td>$state</td> <td>$zip</td> </tr> </table><br /> "; } } else if ($dropdownValue == "city"){ $sql = mysql_query("SELECT * FROM pcparts WHERE city='$searchField'"); while($row1 = mysql_fetch_array($sql)){ $arrayuserinfo[] = array( 'phone' => $row1["phone"], 'name' => $row1["name"], 'city' => $row1["city"], 'state' => $row1["state"], 'address' => $row1["address"], 'zip' => $row1["zip"], ); }; for($i=0;$i < count($arrayuserinfo);$i++){ $phone = $arrayuserinfo[$i]["phone"]; $name = $arrayuserinfo[$i]["name"]; $city = $arrayuserinfo[$i]["city"]; $state = $arrayuserinfo[$i]["state"]; $address = $arrayuserinfo[$i]["address"]; $zip = $arrayuserinfo[$i]["zip"]; echo"<table width='400' border='1' cellpadding='3'> <tr> <td>phone</td> <td>name</td> <td>address</td> <td>city</td> <td>state</td> <td>zip</td> </tr> <tr> <td>$phone</td> <td>$name</td> <td>$address</td> <td>$city</td> <td>$state</td> <td>$zip</td> </tr> </table><br /> "; } } } ?> <html> <body> <h2>Customers Database Search</h2> <form action="file1.php" method="POST"> <input type='hidden' name='parse_var' value='contactform'> Search fields: <select name="dropdown" id="dropdown" > <option value="<?php print "$dropdownValue"; ?>"><?php print "$dropdownValue"; ?></option> <option value="phone">phone</option> <option value="name">name</option> <option value="city">city</option> </select> Search item:<input type="text" name='searchinput' id="searchinput" value=" <?php print "$searchField"; ?>" size='20'><br><br> <input type="submit" name="button" id="button" value="Search" /> </form> <br /> <br /> </body> </html> mysql_connect:
<?php $db_host = "localhost"; $db_username = "user"; $db_pass = "test1234"; $db_name = "pcparts"; @mysql_connect("$db_host","$db_username","$db_pass") or die ("Could not connect to MySQL"); @mysql_select_db("$db_name") or die ("No database"); ?>
если вы найдете что-то, что мне нужно добавить, не могли бы вы рассказать мне, где именно?