I am trying to insert some data into my postgresql database table using an html form and a php script. The problem here is that when the script is run, it does not insert data into the last two coulmns and also it would insert a 0 into the third column.
but when i run the query directly on the database using this sql
INSERT INTO crops (crop_id,crop_type,crop_name,cultivation_yrs, local_name)VALUES('6','food crop','garri','6','utara');
it inserts the data very well meaning that something is wrong with my php script
Please someone should help to tell me what im missing out in my scripts here.
script 1 is my html form script. - <html>
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<head><title>Crop form</title>
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</head>
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<body>
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<p>
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<form action="inputcrops.php" method="post">
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<table width="600" cellpadding="10" cellspacing="0" border="2">
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<tr align="center" valign="top">
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<td align="center" colspan="1" rowspan="1" bgcolor="#64b1ff">
-
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Input crop_id: <input type="text" name="crop_id"><br>
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Input crop_type: <input type="text" name="crop_type"><br>
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Input crop_name: <input type="text" crop_name="crop_name"><br>
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Input cultivation_yrs: <input type="text" cultivation_yrs="cultivation_yrs"><br>
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Input local_name: <input type="text" local_name="local_name"><br>
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<input type="Submit" name="submit" value="Submit">
-
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</form></P>
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</body>
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</html>
script two is my php script - <html>
-
<body>
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<?php
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$PGHOST = localhost;
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$PGDATABASE = "SantaRosaDB";
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$PGUSER = "****";
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$PGPASSWORD = "****";
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$PGPORT = 5432;
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$db_handle = pg_connect("dbname=$PGDATABASE user=$PGUSER
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password=$PGPASSWORD");
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if ($db_handle) {
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echo 'Connection attempt succeeded.';
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} else {
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echo 'Connection attempt failed.';
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}
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$crop_id = intval( $_POST['crop_id']);
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$crop_type = pg_escape_string($_POST['crop_type']);
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$crop_name = pg_escape_string($_POST['crop_name']);
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$cultivation_yrs = intval( $_POST['cultivation_yrs']);
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$local_name = pg_escape_string($_POST['local_name']);
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$query = "INSERT INTO crops (crop_id,crop_type,crop_name,cultivation_yrs, local_name)VALUES($crop_id,$crop_type,$crop_name,$cultivation_yrs,$local_name)";
-
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// INSERT INTO crops (crop_id,crop_type,crop_name,cultivation_yrs, local_name)VALUES('6','food crop','garri','6','utara');
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$result = pg_query($query);
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if (!$result) {
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$errormessage = pg_last_error();
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echo "Error with query: " . $errormessage;
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exit();
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}
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printf ("These values were inserted into the database - %s %s %s %s %s",$crop_id,$crop_type,$crop_name,$cultivation_yrs,$local_name);
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pg_close();
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?>
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</body>
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</html>
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Here is the response i get when i run the code
Connection attempt succeeded.These values were inserted into the database - 1 food crop 0
11  2553 
You need to wrap quotes around string values - $query = "INSERT INTO crops (crop_id,crop_type,crop_name,cultivation_yrs, local_name)
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VALUES($crop_id,'$crop_type','$crop_name',$cultivation_yrs,'$local_name')";
It doesn't seen to do any harm wrapping quotes around numerical values also
I have tried to enclose the string values in quites but it is giving me this error - $query = "INSERT INTO crops (crop_id,crop_type,crop_name,cultivation_yrs, local_name)VALUES('$crop_id', '"$crop_type"', '"$crop_name"','$cultivation_yrs','"$local_name"')";
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$result = pg_query($query);
Parse error: syntax error, unexpected T_VARIABLE in C:\ms4w\Apache\htdocs\php\inputcrops.php on line 21
when i changed it to this i still get the same error - $query = "INSERT INTO crops (crop_id,crop_type,crop_name,cultivation_yrs, local_name)VALUES('$crop_id', "$crop_type", "$crop_name",'$cultivation_yrs',"$local_name")";
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$result = pg_query($query);
Please i really need help on this code, its driving me crazy
thanks
Parse error: syntax error, unexpected T_VARIABLE in C:\ms4w\Apache\htdocs\php\inputcrops.php on line 21
the double quotation marks end the string and PHP expects either a semicolon or a dot operator to extend the string.
when you said that php expects a dot or semi-colon to extend the string what do you mean exactly.
I have these but its still giving me the same error - $query = "INSERT INTO crops (crop_id,crop_type,crop_name,cultivation_yrs,local_name)VALUES('crop_id', "$crop_type"; "$crop_name"; "$cultivation_yrs"; "$local_name")";
- $query = "INSERT INTO crops (crop_id,crop_type,crop_name,cultivation_yrs,local_name)VALUES('crop_id' "$crop_type" "$crop_name" "$cultivation_yrs" "$local_name")";
- $query = "INSERT INTO crops (crop_id,crop_type,crop_name,cultivation_yrs,local_name)VALUES('crop_id',."$crop_type".,."$crop_name".,."$cultivation_yrs".,."$local_name".)";
Parse error: syntax error, unexpected T_VARIABLE in C:\ms4w\Apache\htdocs\php\inputcrops.php on line 21
None of the versions you are showing match the example I gave.
Look at my example and you will understand.
when you said that php expects a dot or semi-colon to extend the string what do you mean exactly.
I’m talking about basic PHP syntax. - $query = "INSERT INTO crops (crop_id,crop_type,crop_name,cultivation_yrs,local_name)VALUES('crop_id', "$crop_type"; "$crop_name"; "$cultivation_yrs"; "$local_name")";
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=> => => ^ end of string
correct usage of quotation marks: - // quotes inside a string
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$str = "bla 'bla' blubb";
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$str = 'bla "bla" blubb';
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$str = "bla \"bla\" blubb";
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$str = 'bla \'bla\' blubb';
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// a variable inside a string
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$str = "bla bla $blubb";
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// concatenate strings (and variables)
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$str = "bla bla" . "blubb";
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$str = "bla bla" . 'blubb';
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$str = "bla bla" . $blubb;
I have tried what i understand from your example but i am still having the same error on the same line 21 . Please could you check to see if i am getting you right. Here is the code.
Thanks - $query = "INSERT INTO crops (crop_id,crop_type,crop_name,cultivation_yrs,local_name)VALUES('crop_id','"$crop_type"','"$crop_name"','"$cultivation_yrs"','"$local_name"')";
(¬_¬)
what is so difficult in understanding that the string goes from one " to the next " (no matter what your intention is)?
there have been plenty of demos, how to use quotation marks correctly.
You have not said anything different from what i did and i am saying, i have done that but it is not giving me any different solution.
Please all i need you to do is tell me if i did not put my double quotes right (which i am thinking i did).
Try and be patient with me, i am not a php expert like you. I am just here to learn from people like you.
Below is what i have done, i am still not right, please be patient and correct me
$query = "INSERT INTO crops (crop_id,crop_type,crop_name,cultivation_yrs,local _name)VALUES('crop_id',"$crop_type","$crop_name"," $cultivation_yrs","$local_name")";
Please all i need you to do is tell me if i did not put my double quotes right (which i am thinking i did).
you did not put your double quotes right. your line contains 10 double quotes (i.e. 5 separate strings) without any string operation inbetween. remove (or escape) them but two (first and last) and the error should vanish. strings in PHP
PS. that’s not expert knowledge, this is something, you need in every programming and markup language
Sorry Dormilich, but line 10 has to do only with my database connection. I dont have any error on line 10. Postgresql does not accept a separator comma or semicolon, it only accepts space in between the strings.
The database connection portion of the script which is from line 4 to line 14 runs perfectly okay and report success in connection. My problem lies on line 21 which i dont quite get.
The script inserts data into only 3 out of the five columns and omits the remaining colums, populating them with just 0 and ".
It looks strange to me.
if you could just let me know if there is any problem observed on my line 21.
I would appreciate it.
Thanks
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