php - Invalid MySQL expression? -

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इस सवाल का पहले से ही एक उत्तर है: < 33 उत्तर

मेरे पास एक निम्न कोड है, जिसे पहचानकर्ता (नाम) के साथ एक पंक्ति मिलनी चाहिए: 

  & lt;? Php शामिल हैं "constants.php"; $ Namee = $ _ पोस्ट [ 'नाम']; $ Passw = $ _ पोस्ट [ 'पासवर्ड']; गूंज $ namee। "& Lt; br & gt;"; प्रतिधारा $ passw। "& Lt; br & gt;"; $ Databaseconnection = mysqli_connect ( "", dbuser, userpassword, "db"); $ Res = mysqli_query ($ डेटाबेस कन्नेक्शन, "चयन पहचानकर्ता, पासवर्ड से खातों में पहचानकर्ता = $ namee"); जबकि ($ पंक्ति = mysqli_fetch_array ($ res)) {print_r ($ पंक्ति); }? & Gt;

मुझे जो त्रुटि मिलती है वह है "चेतावनी: mysqli_fetch_array () पैरामीटर 1 को mysqli_result होने की उम्मीद है, बुलियन C: \ Xampp \ htdocs \ tests \ loginlogic.php पंक्ति 15 पर दिया" जिसका अर्थ है ऐसा कोई रिकॉर्ड नहीं है मुझे लगता है "चयनकर्ता पहचानकर्ता, पासवर्ड से खातों में पहचानकर्ता = $ namee" गलत है। किस तरह से?

आपको $ namee

के आसपास उद्धरण की आवश्यकता है 
  WHERE पहचानकर्ता = '$ namee'

लेकिन आपको वास्तव में अपने यूजर इनपुट से बचने या एक तैयार वक्तव्य का उपयोग करने की आवश्यकता है, ताकि SQL इंजेक्शन।


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